微積分

節錄微積分課程的部分公式與例題

Ch4. Integration 積分

4.1 Antiderivatives & Indefinite Integration

$$ F'(x) = f(x) \;\Rightarrow\; F(x) \text{ is an antiderivative of } f(x) $$

不定積分的表示：
$$ \int f(x)\,dx = F(x) + C $$

\[ \text{Intergration Formula} \]

\begin{align} \int 0 \, dx &=C \\ \int k \, dx &= kx+c \\ \int kf(x) \,dx &= k \int f(x) \, dx \\ \int [f(x)\pm g(x)] \, dx &= \int f(x) \, dx \pm \int g(x) \, dx \\ \int x^n\,dx &= \frac{x^{n+1}}{n+1}+C,\;n\ne -1 \\ \int cos \, x \, dx &= sin \, x+C \\ \int sin \, x \, dx &= -cos \, x+C \\ \int sec^2 \, x \, dx &= tan \, x+C \\ \int cos^2 \, x \, dx &= -cot \, x+C \\ \int sec \, x \, tan \, x \, dx &= sec\,x+C \\ \int csc \, x \, cot \, x \, dx &= -csc \, x+C \end{align}

範例

\( \displaystyle \text{EX:} \quad \int \frac{sin \, x}{cos^{2} \, x} \, dx \)

\begin{align} \int \frac{sin \, x}{cos^{2} \, x} \, dx &=\int(\frac{1}{cos\,x})(\frac{sin\,x}{cos\,x})\,dx \\ &=\int sec\,x\,tan\,x\,dx \\ &=sec\,x+C \end{align}

4.2 Area

\( \displaystyle \sum_{i=1}^{n}a_i=a_1+a_2+a_3+...+a_n \)


\[ \text{Summation Formula} \]

\( \displaystyle \begin{align*} \sum_{i=1}^{n}c &= c\cdot n,\;c\;is\;a\;constant \\ \sum_{i=1}^{n}i &= \frac{n(n+1)}{2} \\ \sum_{i=1}^{n}i^2 &= \frac{n(n+1)(2n+1)}{6} \\ \sum_{i=1}^{n}i^3 &= \frac{n^2(n+1)^2}{4} \end{align*} \)

4.3 黎曼和與定積分 (Riemann Sums & Definite Integrals)

\( \displaystyle \textbf{Thm} \quad \int_{a}^{b} f(x) \, dx = \lim_{\max \Delta x_i \to 0} \sum_{i=1}^{n} f(x_i^{*})\,\Delta x_i \)

4.4 微積分基本定理 (The Fundamental Theorem of Calculus)

\( \textbf{Thm} \quad \text{suppose f is continuous on[a,b]} \)

(Ⅰ)

\[ \begin{align} \text{Let} \quad F(x) &= \int_{a}^{x} f(t)\,dt ,\quad \forall x \in[a,b], \\ \text{Then} \quad F'(x) &= \frac{d}{dx}\int_a^x f(t) \, dt =f(x) \end{align} \]

(Ⅱ)\[ \begin{align} \text{If}& \quad G'(x) = f(x),\quad \forall x\in [a,b],\quad \\ \text{Then}& \quad \int_{a}^{b}f(x)\,dx = G(b)-G(a)=G(x)|_{a}^{b} \end{align} \]

\( \textbf{Pf} \)

(a)
(i) \[ \lim_{h \to 0} \frac{F(x+h)-F(x)}{h}=f(x),\quad let\;x\in[a,b),\;h>0,\;By\;MVT\;for\;intergral \]

$$ \frac{F(x+h)-F(x)}{h}=\frac{\int_{x}^{x+h}f(t)\,dt}{h}=\frac{f(c)\cdot h}{h},\quad where\;c\in\left[x,x+h\right] $$

$$ \lim_{h \to 0^+}\frac{F(x+h)-F(x)}{h}=\lim_{h \to 0^+}f(c)=\lim_{c \to x^+}f(c)=f(x),\quad since\;f\;is\;conti\;at\;x $$

(ii) \[ Let\;x\in(a,b],\;\text{similarly} \quad\lim_{h \to 0^+}\frac{F(x+h)-F(x)}{h}=f(x) \]


(b) \( Suppose\;G'(x)=f(x),\;x\in[a,b] \)

\( By(a),\;F'(x)=f(x),\;x\in[a,b]\;then\;F'(x)=G'(x),\;x\in[a,b] \)

$$ get\;F(x)=G(x)+c $$

$$ 0=\int_{a}^{a}f(t)\,dt=F(a)=G(a)+c,\;c=-G(a) $$

$$ \int_{a}^{b}f(t)\,dt=F(b)=G(b)-G(a),\quad Q.E.D $$


Thm

\( Suppose\;f\;is\;conti\;and\;g,h\;are\;differentiable,\;then \)

(Ⅰ) \( \frac{d}{dx}\int_{a}^{g(x)}f(t)\,dt=f(g(x))\cdot g'(x) \)

(Ⅱ) \( \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)\,dt=f(g(x))\cdot g'(x)-f(h(x))\cdot h'(x) \)

Pf

\( Let\;F(x)=\int f(t)\,dt,\;then \)
\(\frac{d}{dx}\int_{a}^{g(x)}f(t)\,dt=\frac{d}{dx}F(g(x))=F'(g(x))\cdot g'(x)=f(g(x))\cdot g'(x) \)
\( \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)\,dt=\frac{d}{dx}F(t)|_{h(x)}^{g(x)}=\frac{d}{dx}(F(g(x)-F(h(x)))=F'(g(x))\cdot g'(x)-F'(h(x))\cdot h'(x) \)

\( EX.\quad \frac{d}{dx}\int_{x}^{1}\sqrt{1+t^2}\,dt=-\sqrt{1+x^2} \)

\(EX.\quad \frac{d}{dx}\int_{1}^{3x}\frac{sin\,t}{t}\,dt\)

\( Let\;g(x)=3x,\;f(x)=\frac{sin\,x}{x},\;then \)
$$ \frac{d}{dx}\int_{1}^{3x}\frac{sin\,t}{t}\,dt=\frac{sin\,3x}{3x}\cdot 3=\frac{sin\,3x}{x} $$

4.5 代換積分法(Integration by Substitution)

\( \textbf{Thm}\)

\( Suppose\;F'(x)=f(x)\quad F\;is\;an\;antiderivative\;of\;f \)

$$ Then\;\int f(g(x))\cdot g'(x)\,dx=F(g(x))+c $$

\( \textbf{Proof}\)

\( Let\;u=g(x),\;du=dg(x)=g'(x)\,dx \)

$$ \int f(g(x))\cdot g'(x)\,dx=\int f(u)\,du=F(u)+c=F(g(x))+c $$

\( \textbf{EX.} \quad \int (x-1)^4\cdot 2x\,dx \)

$$ Let\;u=x^2-1,\;du=2x\,dx $$ \[ \begin{align} \text{Then} \quad \int (x-1)^4\cdot 2x\,dx &=\int u^4\,du \\ &= \frac{u^5}{5}+c \\ &=\frac{(x^2-1)^5}{5}+c \end{align} \]

5.1 The Natural Logarithmic Function: Differentiation

\( \textbf{Thm}\) \( ln \; x= \int_1^x \frac{1}{t} \, dt, \quad x > 0 \quad \text{is called the natural logarithmic function.} \)

\( \textbf{Proof} \)

\( \text{Since} \; \frac{1}{t} \text{ is conti on } (0, \infty) , \quad ln \, t \; \text{ is defined on } (0, \infty) \)
$$ \frac{d}{dx} \ln x = \frac{d}{dx} \int_1^x \frac{1}{t} \, dt = \frac{1}{x} > 0, \quad \forall x > 0 $$ $$ \frac{d^2}{dx^2} \ln x = -\frac{1}{x^2} < 0, \quad \forall x>0 $$


$$ \text{The domain of} \; ln \; \text{is} (0, \infty), \; \text{range of} \; ln \; \text{is} (-\infty, \infty) $$

$$ ln(x) = \begin{cases} > 0, & \text{if } x>1 \\ 0, & \text{if } x = 1 \\ < 0, & \text{if } x < 1 \end{cases} \quad \ln \text{ is conti, increasing, one-to-one} $$

\( \textbf{Thm} \quad \frac{d}{dx} \ln \, x = \frac{1}{x}, \quad \forall x > 0, \quad \int \frac{1}{x} \, dx = \ln|x| + C, \quad x \neq 0 \)

\( \textbf{Thm} \quad a, \, b>0 \quad r\in Q \)

\[ ln(a \cdot b) = ln a + ln b \]\[\ln a^r = r \ln a \]\[ \ln \, \frac{a}{b} = \ln a - \ln b \]

\( \textbf{EX.} \quad f(x) = \ln \sqrt{x^2 + 1} \) \[ \begin{align} f'(x) &= \frac{1}{\sqrt{x^2+1}} \cdot \frac{d}{dx}\sqrt{x^2+1} \\ &= \frac{1}{\sqrt{x^2+1}} \cdot \frac{1}{2\sqrt{x^2+1}} \cdot 2x \\ &= \frac{x}{x^2 + 1} \end{align} \]


\( \displaystyle \textbf{EX.} \quad f(x) = \ln\left(\frac{x(x^2+1)^2}{\sqrt{2x^5 -1}}\right) \) \[ \begin{align} f(x) &= \ln\left(\frac{x(x^2+1)^2}{\sqrt{2x^5 -1}}\right) \\ &= \ln x + 2\ln(x^2+1) - \ln\sqrt{2x^5 -1} \\ f'(x) &= \frac{1}{x} + \frac{2}{x^2+1} \cdot 2x - \frac{1}{2(2x^5 -1)} \cdot 10x^4 \\ &= \frac{1}{x} + \frac{4x}{x^2+1} - \frac{5x^4}{2x^5 -1} \end{align} \]

5.2 The Natural Logarithmic Function: Integration

\( \displaystyle \textbf{Thm} \quad \frac{d}{dx}\ln|x| = \frac{1}{x}, \quad x \neq 0 \)

\( \textbf{Pf} \) \[ \frac{d}{dx}\ln|x| = \begin{cases} \frac{d}{dx}\ln x = \frac{1}{x}, \quad x>0 \\ \frac{d}{dx}\ln (-x) = -\frac{1}{x} \frac{d}{dx} (-x)=\frac{1}{x} , \quad x<0 \end{cases} \] \[ \text{s.t.} \quad \int \frac{1}{u}\,du = \ln|u| + C \] \[ \text{Cor:If　} f \text{ is diff and } f(x)\neq 0, \text{ then } \frac{d}{dx}\ln|f(x)| = \frac{f'(x)}{f(x)} \]

\( \textbf{EX.} \quad f(x) = \ln|\sec x| \) \[ \begin{align} f'(x) &= \frac{1}{\sec x} \cdot \sec x \tan x \\ &= \tan x \\ \Rightarrow \int \tan x\,dx &= \ln|\sec x| + C \end{align} \]

\( \textbf{EX.} \quad \int \cot x\,dx = \ln|\sin x| + C \)

\[ \begin{align} \int \sec x\,dx &= \int \frac{\sec x(\sec x + \tan x)}{\sec x + \tan x} dx \\ &= \int \frac{du}{u} \\ &= \ln|u| + C \\ &= \ln|\sec x + \tan x| + C \end{align} \]

$$ \int \csc x\,dx = \int \frac{\csc x(\csc x - \cot x)}{\csc x - \cot x} dx $$ $$ = \int \frac{du}{u} = \ln|u| + C $$ $$ = \ln|\csc x - \cot x| + C $$

\( \textbf{EX.} \)

\[ \begin{align*} \int \frac{x}{x^2+1} \, dx &= \int \frac{1}{2u} \, du \quad \left( \text{Let } \begin{cases} u = x^2+1 \\ du = 2x \, dx \end{cases} \right) \\[6pt] &= \frac{1}{2} \ln|x^2+1| + C \end{align*} \]

\[ \begin{align*} \int \frac{2x^2}{x-1} \, dx &= \int \left( 2x + 2 + \frac{2}{x-1} \right) dx \\[6pt] &= x^2 + 2x + 2 \ln|x-1| + C \end{align*} \]

\[ \begin{align*} \int \frac{1}{x \ln x} \, dx &= \int \frac{1}{\ln x} \cdot \frac{1}{x} \, dx \quad \left( \text{Let } \begin{cases} u = \ln x \\ du = \frac{1}{x} \, dx \end{cases} \right) \\[6pt] &= \int \frac{du}{u} \\[6pt] &= \ln |\ln x| + C \end{align*} \]

\( \textbf{EX.} \displaystyle \quad \text{Find } \frac{dy}{dx} \; \text{for } y = \frac{x^{3/4} \sqrt{x^2+1}}{(3x+5)^5} \)

\[ \Rightarrow\quad \ln y = \frac{3}{4} \ln x + \frac{1}{2} \ln(x^2+1) - 5 \ln(3x+5) \] \[ \begin{align*} \frac{d}{dx} (\ln y) &= \frac{1}{y}\frac{dy}{dx} \\[6pt] &= \frac{3}{4x} + \frac{x}{x^2+1} - \frac{15}{3x+5} \\[6pt] \Rightarrow\quad \frac{dy}{dx} &= y\left( \frac{3}{4x} + \frac{x}{x^2+1} - \frac{15}{3x+5} \right) \end{align*} \]

5.3 Inverse function

\( \textbf{Def.} \)

\[ \text{When} \displaystyle \begin{cases} f \left( g(x) \right) = x \\ g\left(f(x) \right)x \end{cases} \] \[ \text{s.t.} \quad g(x)=f^{-1}(x) \]

Derivative of an Inverse Function

\( \textbf{Thm} \)

\( \text{Let } f \text{ be a function that is differentiable on an interval } I \)

\( \text{ If } f \text{ has an inverse function } g \text{, then } g \text{ is differentiable at any } x \text{ for which } f'(g(x)) \ne 0 \)

\[ g'(x) = \frac{1}{f'(g(x))}, \quad f'(g(x)) \ne 0 \]

\( \textbf{Pf}\)

\[ f(g(x)) = x \] \[ \begin{align*} \frac{d}{dx}[f(g(x))] &= \frac{d}{dx}[x] \\[6pt] f'(g(x)) \cdot g'(x) &= 1 \end{align*} \] \[ g'(x) = \frac{1}{f'(g(x))} \]

5.4 Exponential Function

\( \textbf{Def.} \quad f(x)= ln \, x ,\; f^{-1}(x) = e^{x} \)

\( \textbf{Thm.} \quad \text{Let } a \text{ and } b \text{ be any real numbers.} \)

\[e^a e^b = e^{a+b} \] \[ \frac{e^a}{e^b} = e^{a-b} \]

\( \textbf{Pf.} \)

\[ \displaystyle \begin{align*} \ln(e^a e^b) &= \ln(e^a) + \ln(e^b) \\ &= a + b \\ &= \ln(e^{a+b}) \end{align*} \]

\( \text{Since } \ln x \text{ is one-to-one, we conclude that } e^a e^b = e^{a+b} \text{.} \)

Differentiation and Integration of Exponential Functions

\( \textbf{Thm.} \quad \text{Let } u \text{ be a differentiable function of } x \text{.} \)

\[\frac{d}{dx}[e^x] = e^x \]\[ \frac{d}{dx}[e^u] = e^u \frac{du}{dx}\]

\( \textbf{Pf.} \quad \text{Let } y = e^x \text{, then } \ln y = x \text{.} \)

\[ \displaystyle \begin{align*} \ln y &= x \\[6pt] \frac{d}{dx}[\ln y] &= \frac{d}{dx}[x] \quad \\[6pt] \frac{1}{y} \frac{dy}{dx} &= 1 \quad \\[6pt] \frac{dy}{dx} &= y \quad \\[6pt] \frac{d}{dx}[e^x] &= e^x \end{align*} \]

\( \text{The derivative of } e^u \text{ follows from the Chain Rule.} \)

\( \textbf{Thm.} \quad \text{Let } u \text{ be a differentiable function of } x \)

\[\int e^x \, dx = e^x + C \]\[\int e^u \, du = e^u + C \]

\( \displaystyle \textbf{Ex} \quad \text{Find } \int e^{3x+1}\,dx \text{.} \)

\( \text{let } u = 3x + 1, \text{ then } du = 3\,dx \)

\[ \displaystyle \begin{align*} \Rightarrow \int e^{3x+1}\,dx &= \frac{1}{3} \int e^{3x+1}(3)\,dx \\[6pt] &= \frac{1}{3} \int e^u\,du \\[6pt] &= \frac{1}{3} e^u + C \\[6pt] &= \frac{1}{3} e^{3x+1} + C\end{align*} \]

\( \displaystyle \textbf{Ex.} \quad \text{Find }\int 5x e^{-x^2}\,dx \)

\( \displaystyle \text{let } u = -x^2, \text{ then } du = -2x\,dx \quad x\,dx = \frac{-du}{2}\)

\[ \displaystyle \begin{align*} \int 5x e^{-x^2} dx &= \int 5 e^{-x^2} (x dx) \\[6pt] &= \int 5 e^u \left( -\frac{du}{2} \right) \\[6pt] &= -\frac{5}{2} \int e^u du\\[6pt] &= -\frac{5}{2} e^u + C \\[6pt] &= -\frac{5}{2} e^{-x^2} + C \end{align*} \]

\( \displaystyle \textbf{Ex.} \quad \text{Find the indefinite integral } \int \frac{e^{\frac{1}{x}}}{x^2} \, dx \)

\( \displaystyle \text{Let } u = \frac{1}{x}, \text{ then } du = \frac{-1}{x^{2}} \,dx \)

\[ \displaystyle \begin{align*} \Rightarrow \int \frac{e^{\frac{1}{x}}}{x^2} dx &= \int e^{\frac{1}{x}} \left( \frac{1}{x^2} dx \right)\\[6pt] &= \int e^u\,(-du)\\[6pt] &= - \int e^u \, du \\[6pt] &= -e^u + C \\[6pt] &= -e^{1/x} + C \end{align*} \]

\( \textbf{Ex.} \quad \displaystyle \text{Evaluate the definite integral } \int_0^1 \frac{e^x}{1+e^x} \,dx \)

\( \text{Let } u = 1+e^x, \text{ then } du = e^x \,dx \text{.} \)

\[ \displaystyle \begin{align*} \Rightarrow \int_0^1 \frac{e^x}{1+e^x} \,dx &= \int_{x=0}^{x=1} \frac{1}{1+e^x} (e^x \,dx) \\[6pt] &= \int_{u=1+e^0}^{u=1+e^1} \frac{1}{u} \,du \\[6pt] &= \int_2^{1+e} \frac{1}{u} \,du \\[6pt] &= \left[ \ln|u| \right] |_2^{1+e} \\[6pt] &= \ln(1+e) - \ln(2) \\[6pt] &= \ln\left(\frac{1+e}{2}\right) \\[6pt] &\approx 0.620 \end{align*} \]

5.5 Bases Other than e and Applications

\( \textbf{Def.} \quad \displaystyle \text{If } a \text{ is a postive real number} (a \ne 1) \text{, and } x \in \mathbb{R} \)

\[ a^{x} = e^{(ln \, a)x} \] \[ log_{a} \, x = \frac{1}{ln \, a} ln \, x \]

\( \textbf{Thm} \quad \text{Derivatives for Bases Other than } e \)

\( \begin{align*} &\text{Let } a \text{ be a positive real number } (a \neq 1) \\[6pt] &\text{ and let } u \text{ be a differentiable function of } x \text{.} \end{align*} \)

\[ \frac{d}{dx}[a^x] = (\ln a) a^x \] \[ \frac{d}{dx}[a^u] = (\ln a) a^u \frac{du}{dx} \] \[ \frac{d}{dx}[\log_a x] = \frac{1}{(\ln a) x} \] \[ \frac{d}{dx}[\log_a u] = \frac{1}{(\ln a) u} \frac{du}{dx} \]

\( \textbf{Pf.} \)

\[ \frac{d}{dx}[a^x] = \frac{d}{dx}[e^{(\ln a)x}] = e^{(\ln a)x} (\ln a) = a^x (\ln a) \]

\[ \begin{align*} \frac{d}{dx}[\log_a x] &= \frac{d}{dx}\left[\frac{\ln x}{\ln a}\right] = \frac{1}{\ln a} \frac{d}{dx}[\ln x] \\[6pt] &= \frac{1}{\ln a} \left(\frac{1}{x}\right) = \frac{1}{(\ln a)x} \end{align*} \]

\( \textbf{Ex} \quad \text{Find the derivative of each function} \)

\( \displaystyle \textbf{a.} \quad y = 2^x \)

\[ \displaystyle y' = \frac{d}{dx}[2^x] = (\ln 2) 2^x \]

\( \displaystyle \textbf{b.} \quad y = 2^{3x-1} \)

\[ \displaystyle \begin{align*} y' &= \frac{d}{dx}[2^{3x-1}] = (\ln 2) 2^{3x-1} \frac{d}{dx}[3x-1] \\[6pt] &= (\ln 2) 2^{3x-1} (3) = 3(\ln 2) 2^{3x-1} \end{align*} \]

\( \displaystyle \textbf{c.} \quad y = \log_{10} (\cos x) \)

\[ \begin{align*} y' &= \frac{d}{dx}[\log_{10} (\cos x)] = \frac{1}{(\ln 10) (\cos x)} \frac{d}{dx}[\cos x] \\[6pt] &= \frac{-\sin x}{(\ln 10) (\cos x)} = -\frac{1}{\ln 10} \tan x \end{align*} \]

\( \textbf{d.} \quad y = \log_3 \frac{\sqrt{x}}{2x+5} \)

\[ y = \log_3 \frac{\sqrt{x}}{2x+5} = \frac{1}{2} \log_3 x - \log_3 (2x+5) \] \[ \displaystyle \begin{align*} y' &= \frac{d}{dx}\left[\frac{1}{2} \log_3 x - \log_3 (2x+5)\right] \\[6pt] &= \frac{1}{2} \left( \frac{1}{(\ln 3) x} \right) - \frac{1}{(\ln 3) (2x+5)} \frac{d}{dx}[2x+5] \\[6pt] &= \frac{1}{2(\ln 3) x} - \frac{1}{(\ln 3) (2x+5)} (2) \\[6pt] &= \frac{1}{2(\ln 3) x} - \frac{2}{(\ln 3) (2x+5)} \\[6pt] &= \frac{(2x+5) - 2(2x)}{2(\ln 3) x (2x+5)} \\[6pt] &= \frac{2x+5 - 4x}{2(\ln 3) x (2x+5)} \\[6pt] &= \frac{5 - 2x}{2(\ln 3) x (2x+5)} \end{align*} \]

\( \textbf{Thm} \quad \text{Let } n \in \mathbb{R} \text{ , and let } u \text{ be a differentiable function of } x \text{.} \)

\[ \frac{d}{dx}[x^n] = nx^{n-1} \] \[ \frac{d}{dx}[u^n] = nu^{n-1} \frac{du}{dx} \]

\( \displaystyle \textbf{EX.} \quad \text{Find the derivative of each function.} \)

\( \textbf{a.} \quad \frac{d}{dx}[e^e] = 0 \)

\( \textbf{b.} \quad \frac{d}{dx}[e^x] = e^x \)

\( \textbf{c.} \quad \frac{d}{dx}[x^e] = ex^{e-1} \)

\( \textbf{d.} \quad y = x^x \)

\[ \displaystyle \begin{align*} y &= x^x \\[6pt] \ln y &= \ln x^x \\[6pt] \ln y &= x \ln x \\[6pt] \frac{y'}{y} &= x \left(\frac{1}{x}\right) + (\ln x)(1) \\[6pt] \frac{y'}{y} &= 1 + \ln x \\[6pt] y' &= y(1 + \ln x) \\[6pt] y' &= x^x(1 + \ln x) \end{align*} \]

5.6 Indeterminate Forms and L'Hôpital's Rule

\( \textbf{Thm} \quad \text{The Extended Mean Value Theorem} \)

\( \begin{align*} &\text{If } f \text{ and } g \text{ are differentiable on an open interval } (a, b) \text{ and continuous on } [a, b] \\ &\text{ such that } g'(x) \neq 0 \text{ for any } x \text{ in } (a, b) \text{, then there exists a point } c \text{ in } (a, b) \end{align*} \)

\[ \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} \]

\( \text{A proof of this theorem is given in Appendix A.} \)

\( \textbf{Thm} \quad \text{L'Hôpital's Rule} \)

\( \begin{align*} &\text{Let } f \text{ and } g \text{ be functions that are differentiable on an open interval } (a, b) \text{ containing } c \\ &\text{, except possibly at } c \text{ itself. If the limit of } f(x)/g(x) \text{ as } x \text{ approaches } c \text{ produces the indeterminate form } 0/0 \text{, then} \end{align*} \)

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

\( \begin{align*} &\text{provided the limit on the right exists (or is infinite). } \\ &\text{This result also applies when the limit of } f(x)/g(x) \text{ as } x \text{ approaches } c \text{ produces one of the indeterminate forms } \infty/\infty \text{, } (-\infty)/\infty \text{, } \infty/(-\infty) \text{, or } (-\infty)/(-\infty) \text{.} \end{align*} \)

\( \text{A proof of this theorem is given in Appendix A.} \)

\( \displaystyle \textbf{EX.} \quad \text{Evaluate the limit } \lim_{x \to 0} \frac{e^{2x} - 1}{x} \text{.} \)

\[ \lim_{x \to 0} (e^{2x} - 1) = e^0 - 1 = 0 \] \[ \lim_{x \to 0} x = 0 \]

\( \text{Apply L'Hôpital's Rule:} \)

\[ \displaystyle \lim_{x \to 0} \frac{e^{2x} - 1}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}[e^{2x} - 1]}{\frac{d}{dx}[x]} = \lim_{x \to 0} \frac{2e^{2x}}{1} = 2e^0 = 2 \]

\( \displaystyle \textbf{EX.} \quad \text{Evaluate the limit } \lim_{x \to \infty} \frac{\ln x}{x} \)

\[ \displaystyle \lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{\frac{d}{dx}[\ln x]}{\frac{d}{dx}[x]} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0 \]

\( \displaystyle \textbf{Ex.} \quad \text{Evaluate the limit } \lim_{x \to -\infty} \frac{x^2}{e^{-x}} \)

\( \text{Apply L'Hôpital's Rule (1st time):} \)

\[ \displaystyle \lim_{x \to -\infty} \frac{x^2}{e^{-x}} = \lim_{x \to -\infty} \frac{\frac{d}{dx}[x^2]}{\frac{d}{dx}[e^{-x}]} = \lim_{x \to -\infty} \frac{2x}{-e^{-x}} \]

\( \text{Apply L'Hôpital's Rule again (2nd time):} \)

\[ \displaystyle \lim_{x \to -\infty} \frac{2x}{-e^{-x}} = \lim_{x \to -\infty} \frac{\frac{d}{dx}[2x]}{\frac{d}{dx}[-e^{-x}]} = \lim_{x \to -\infty} \frac{2}{e^{-x}} \] \[ \lim_{x \to -\infty} \frac{2}{e^{-x}} = \frac{2}{\infty} = 0 \]

\( \displaystyle \textbf{Ex.} \quad \text{Evaluate the limit } \lim_{x \to \infty} e^{-x} \sqrt{x} \)

\( \textbf{Solution} \)

\( \text{Rewrite the limit as the indeterminate form } \infty/\infty \)

\[ \displaystyle \lim_{x \to \infty} e^{-x} \sqrt{x} = \lim_{x \to \infty} \frac{\sqrt{x}}{e^{x}} \]

\( \text{Apply L'Hôpital's Rule:} \)

\[ \displaystyle \lim_{x \to \infty} \frac{\sqrt{x}}{e^{x}} = \lim_{x \to \infty} \frac{\frac{d}{dx}[\sqrt{x}]}{\frac{d}{dx}[e^{x}]} \]

\( \text{Calculate the derivatives:} \)

\[ \frac{d}{dx}[\sqrt{x}] = \frac{d}{dx}[x^{1/2}] = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \] \[ \frac{d}{dx}[e^{x}] = e^{x} \]

\( \text{Substitute back into the limit expression:} \)

\[ \displaystyle \lim_{x \to \infty} \frac{\frac{1}{2\sqrt{x}}}{e^{x}} = \lim_{x \to \infty} \frac{1}{2\sqrt{x} e^{x}} \]\[ \lim_{x \to \infty} \frac{1}{2\sqrt{x} e^{x}} = \frac{1}{\infty} = 0 \]

5.7 Inverse Trigonometric Functions: Differentiation

Definitions of inverse trigonometric functions

Function Domain Range
\[ y=arcsin \, x \text{ ,if } sin \, y = x \] \[ -1 \le x \le 1 \] \[ -\frac{\pi}{2} \le y \le \frac{\pi}{2} \]
\[ y=arccos \, x \text{ ,if } cos \, y = x \] \[ -1 \le x \le 1 \] \[ 0 \le y \le \pi \]
\[ y=arctan \, x \text{ ,if } tan \, y = x \] \[ -\infty \le x \le \infty \] \[ -\frac{\pi}{2} \lt y \lt \frac{\pi}{2} \]
\[ y=arccot \, x \text{ ,if } cot \, y = x \] \[ -\infty \lt x \lt \infty \] \[ 0 \le y \le \pi \]
\[ y=arcsec \, x \text{ ,if } sec \, y = x \] \[ |x| \ge 1 \] \[ 0 \le y \le \pi , y\ne \frac{\pi}{2} \]
\[ y=arccsc \, x \text{ ,if } csc \, y = x \] \[ |x| \ge 1 \] \[ -\frac{\pi}{2} \le y \le \frac{\pi}{2}, y \ne 0 \]

\( \text{If } \quad x \in \text{Domain, } \quad y \in \text{Range ,} \; \text{then} \)

\[ \begin{align*} sin(arcsin \, x) &= x \quad \text{and} \quad arcsin(sin \, y) = y \\[6pt] tan(arctan \, x) &= x \quad \text{and} \quad arctan(tan \, y) = y \\[6pt] sec(arcsec \, x) &= x \quad \text{and} \quad arcsec(sec \, y) = y \end{align*} \]

\( \displaystyle \textbf{EX.} \quad \text{Solve the equation } \arctan(2x-3) = \frac{\pi}{4} \)

\[ \displaystyle \begin{align*} \arctan(2x-3) &= \frac{\pi}{4} \\[6pt] \tan[\arctan(2x-3)] &= \tan\left(\frac{\pi}{4}\right) \\[6pt] 2x - 3 &= 1 \\[6pt] 2x &= 4 \\[6pt] x &= 2 \end{align*} \]

\( \textbf{Thm} \quad \text{Let } u \text{ be a differentiable function of } x \)

\[ \begin{align*} \frac{d}{dx}[\arcsin u] &= \frac{u'}{\sqrt{1 - u^2}} \\[6pt] \frac{d}{dx}[\arccos u] &= \frac{-u'}{\sqrt{1 - u^2}} \\[6pt] \frac{d}{dx}[\arctan u] &= \frac{u'}{1 + u^2} \\[6pt] \frac{d}{dx}[\text{arccot } u] &= \frac{-u'}{1 + u^2} \\[6pt] \frac{d}{dx}[\text{arcsec } u] &= \frac{u'}{|u|\sqrt{u^2 - 1}} \\[6pt] \frac{d}{dx}[\text{arccsc } u] &= \frac{-u'}{|u|\sqrt{u^2 - 1}} \end{align*} \]

\( \textbf{Pf.} \)

\( \text{Let } y = \arcsin x \text{, then } \sin y = x \text{, where } -\frac{\pi}{2} \le y \le \frac{\pi}{2} \)

\[ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{\frac{d}{dy}[\sin y]} = \frac{1}{\cos y} \] \[ \frac{d}{dx}[\arcsin x] = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^2 y}} \] \[ \frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}} \]

\( \text{Applying the Chain Rule} \)

\[ \frac{d}{dx}[\arcsin u] = \frac{u'}{\sqrt{1 - u^2}} \]

\( \textbf{EX.} \quad \text{Find the derivative of each function.} \)

\( \textbf{a.} \quad y = \arcsin(2x) \)

\[ \displaystyle \frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1 - (2x)^2}} = \frac{2}{\sqrt{1 - 4x^2}} \]

\( \textbf{b.} \quad y = \arctan(3x) \)

\[ \displaystyle \frac{d}{dx}[\arctan(3x)] = \frac{3}{1 + (3x)^2} = \frac{3}{1 + 9x^2} \]

\( \textbf{c.} \quad y = \arcsin \sqrt{x} \)

\( \displaystyle \text{Let } u = \sqrt{x} = x^{\frac{1}{2}} \text{, so } u' = \frac{1}{2} x^{-\frac{1}{2}} \text{.} \)

\[ \displaystyle \frac{d}{dx}[\arcsin \sqrt{x}] = \frac{\frac{1}{2} x^{\frac{1}{2}}}{\sqrt{1 - (\sqrt{x})^2}} = \frac{1}{2\sqrt{x}\sqrt{1 - x}} = \frac{1}{2\sqrt{x - x^2}} \]

\( \textbf{d.} \quad y = \text{arcsec } e^{2x} \)

\( \text{Let } u = e^{2x} \text{, so } u' = 2e^{2x} \text{.} \)

\[ \displaystyle \frac{d}{dx}[\text{arcsec } e^{2x}] = \frac{2e^{2x}}{|e^{2x}|\sqrt{(e^{2x})^2 - 1}} = \frac{2e^{2x}}{e^{2x}\sqrt{e^{4x} - 1}} = \frac{2}{\sqrt{e^{4x} - 1}} \]

\( \textbf{Thm} \)

  • \[ \text{If } -1 \le x \le 1 \text{, then } \arcsin x + \arccos x = \frac{\pi}{2} \]
  • \[ \text{If } x \in \mathbb{R} \text{, then } \arctan x + \text{arccot } x = \frac{\pi}{2} \]
  • \[ \text{If } |x| \ge 1 \text{, then } \text{arcsec } x + \text{arccsc } x = \frac{\pi}{2} \]

\( \textbf{Proof} \)

\( \text{Let } f(x) = \arcsin x + \arccos x \)

\[ \begin{align*} \Rightarrow f'(x) &= \frac{d}{dx}[\arcsin x] + \frac{d}{dx}[\arccos x] \\[6pt] &= \left( \frac{1}{\sqrt{1 - x^2}} \right) + \left( \frac{-1}{\sqrt{1 - x^2}} \right) \\[6pt] &= 0 \\[6pt] \Rightarrow f(x) &= \arcsin x + \arccos x = C \end{align*} \]

\( \text{Let } x=0\)

\[ \displaystyle C = \arcsin(0) + \arccos(0) =\frac{\pi}{2} \]

\( \text{s.t. } \arcsin x + \arccos x = \frac{\pi}{2} \text{ for } -1 \le x \le 1\)

\( \textbf{Thm} \)

\[ \text{If } |x| \ge 1 \text{, then } \text{arcsec } x = \cos^{-1}\left(\frac{1}{x}\right) \] \[ \text{If } |x| \ge 1 \text{, then } \text{arccsc } x = \sin^{-1}\left(\frac{1}{x}\right) \]

\( \text{The identity for } \text{arccot } x \text{ is piecewise defined:} \)

\[ \text{arccot } x = \begin{cases} \arctan\left(\frac{1}{x}\right), & x > 0 \\ \pi + \arctan\left(\frac{1}{x}\right), & x < 0 \\ \frac{\pi}{2}, & x = 0 \end{cases} \]

\( \textbf{Pf} \)

\( \text{Let } \theta = \text{arcsec } x \text{.} \text{} \)

\[ x = \sec \theta , \quad \frac{1}{x} = \cos \theta \] \[ \cos^{-1}\left(\frac{1}{x}\right) = \theta \text{} \] \[ \sec^{-1}\, x = \cos^{-1}\left(\frac{1}{x}\right) \]

5.8 Inverse Trigonometric Function: Intergration

Integrals involving inverse trigonometric functions

\( \textbf{Thm} \)

\( \text{Let } u \text{ be a differentiable function of } x \text{, and let } a > 0 \text{.} \text{} \)

\[ \displaystyle \begin{align*} \int \frac{du}{\sqrt{a^2 - u^2}} &= \sin^{-1} \frac{u}{a} + C \\[6pt] \int \frac{du}{a^2 + u^2} &= \frac{1}{a} \tan^{-1} \frac{u}{a} + C \\[6pt] \int \frac{du}{u\sqrt{u^2 - a^2}} &= \frac{1}{a} \sec^{-1} \frac{|u|}{a} + C \end{align*} \]

\( \displaystyle \textbf{Ex} \quad \text{Evaluate } \int \frac{dx}{4x^2+4x+2} \)

\(\text{Let } u = 2x + 1 ,\quad du = 2\,dx ,\quad dx = \frac{1}{2}\,du \)
\[ \displaystyle \begin{align*} \int \frac{dx}{4x^2+4x+2} &=\int \frac{dx}{1 + (2x + 1)^2} \\[6pt] &= \int \frac{\frac{1}{2}\,du}{1^2 + u^2} \\[6pt] &= \frac{1}{2} \int \frac{du}{1^2 + u^2} \\[6pt] &= \frac{1}{2} \left( \frac{1}{1} \tan^{-1} \frac{u}{1} \right) + C \\[6pt] &= \frac{1}{2} \tan^{-1} u + C \end{align*} \] \[ \Rightarrow \int \frac{dx}{4x^2+4x+2} = \frac{1}{2} \tan^{-1}(2x + 1) + C \]

\( \displaystyle \textbf{Ex} \quad \text{Evaluate } \int \frac{dx}{\sqrt{e^{2x}-1}} \)

\[ \displaystyle \text{Let } u = e^x ,\quad du = e^x \,dx ,\quad dx = \frac{du}{e^x} = \frac{du}{u} \]
\[ \displaystyle \int \frac{dx}{\sqrt{e^{2x}-1}} = \int \frac{du/u}{\sqrt{u^2 - 1^2}} = \int \frac{du}{u\sqrt{u^2 - 1}} \text{} \] \[ \displaystyle \int \frac{du}{u\sqrt{u^2 - 1}} = \frac{1}{1} \sec^{-1} \frac{|u|}{1} + C = \sec^{-1} |u| + C \] \[ \RIghtarrow \int \frac{dx}{\sqrt{e^{2x}-1}} = \sec^{-1} e^{x} + C \]

\( \displaystyle \textbf{Ex} \quad \text{Evaluate } \int \frac{dx}{x\sqrt{x^2-16}} \)

\( \text{Let } u = x \text{, so } du = dx \text{.} \quad a^2 = 16 \text{, so } a = 4 \text{.} \)

\[ \begin{align*} \displaystyle \int \frac{dx}{x\sqrt{x^2-16}} &= \frac{1}{4} \text{arcsec } \frac{|x|}{4} + C \\[6pt] &= \frac{1}{4} \cos^{-1}\left(\frac{4}{|x|}\right) + C \\[6pt] &= \frac{1}{4} \cot^{-1} \, \frac{4}{\sqrt{x^2 - 16}} + C \end{align*} \]

5.9 Hyperbolic Functions

Definitions of the Hyperbolic Functions

\[ \begin{align*} \sinh x &= \frac{e^{x} - e^{-x}}{2} \\[6pt] \cosh x &= \frac{e^{x} + e^{-x}}{2} \\[6pt] \tanh x &= \frac{\sinh x}{\cosh x} = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} \\[6pt] \text{csch } x &= \frac{1}{\sinh x}, \quad x \ne 0 \\[6pt] \text{sech } x &= \frac{1}{\cosh x} \\[6pt] \coth x &= \frac{1}{\tanh x}, \quad x \ne 0 \end{align*} \]

\( \textbf{Thm} \)

\[ \cosh^2 x - \sinh^2 x = 1 \] \[ \tanh^2 x + \text{sech}^2 x = 1 \] \[ \coth^2 x - \text{csch}^2 x = 1 \]

\( \textbf{Pf} \)

\[ \begin{align*} \displaystyle &\cosh^2 x - \sinh^2 x \\[6pt] &= \left(\frac{e^{x} + e^{-x}}{2}\right)^2 - \left(\frac{e^{x} - e^{-x}}{2}\right)^2 \\[6pt] \displaystyle \quad &= \frac{e^{2x} + 2e^{x}e^{-x} + e^{-2x}}{4} - \frac{e^{2x} - 2e^{x}e^{-x} + e^{-2x}}{4} \\[6pt] \displaystyle \quad &= \frac{e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}}{4} \\[6pt] \displaystyle \quad &= \frac{4}{4} = 1 \end{align*} \]

Sum and Difference Formulas

\[ \sinh(x + y) = \sinh x \cosh y + \cosh x \sinh y \] \[ \sinh(x - y) = \sinh x \cosh y - \cosh x \sinh y \] \[ \cosh(x + y) = \cosh x \cosh y + \sinh x \sinh y \] \[ \cosh(x - y) = \cosh x \cosh y - \sinh x \sinh y \]

Double-Angle Formulas

\[ \sinh^2 x = \frac{-1 + \cosh 2x}{2} \]\[ \cosh^2 x = \frac{1 + \cosh 2x}{2} \]\[ \sinh 2x = 2 \sinh x \cosh x \] \[ \begin{align*} \cosh 2x &= \cosh^2 x + \sinh^2 x \\[6pt] &= 2\cosh^2 x - 1 \\[6pt] &= 1 + 2\sinh x^{2} x \end{align*} \]

Derivatives and Integrals of Hyperbolic Functions

\( \textbf{Thm} \)

\( \text{Let } u \text{ be a differentiable function of } x \text{.} \text{} \)

\[ \begin{align*} \frac{d}{dx}[\sinh u] &= (\cosh u)u' \\[6pt] \frac{d}{dx}[\cosh u] &= (\sinh u)u' \\[6pt] \frac{d}{dx}[\tanh u] &= (\text{sech}^2 u)u' \\[6pt] \frac{d}{dx}[\coth u] &= -(\text{csch}^2 u)u' \\[6pt] \frac{d}{dx}[\text{sech } u] &= -(\text{sech } u \tanh u)u' \\[6pt] \frac{d}{dx}[\text{csch } u] &= -(\text{csch } u \coth u)u' \end{align*} \]

\( \textbf{Pf}\)

\[ \begin{align*} \frac{d}{dx}[\sinh x] &= \frac{d}{dx}\left[\frac{e^x - e^{-x}}{2}\right] \\[6pt] &= \frac{1}{2} \left( \frac{d}{dx}[e^x] - \frac{d}{dx}[e^{-x}] \right) \\[6pt] &= \frac{1}{2} \left( e^x - (-e^{-x}) \right) \\[6pt] &= \frac{e^x + e^{-x}}{2} \quad \\[6pt] &= \cosh x \end{align*} \]

\( \textbf{Pf}\)

\[ \begin{align*} \frac{d}{dx}[\tanh x] &= \frac{d}{dx}\left[\frac{\sinh x}{\cosh x}\right] \\[6pt] &= \frac{(\cosh x) \frac{d}{dx}[\sinh x] - (\sinh x) \frac{d}{dx}[\cosh x]}{\cosh^2 x} \\[6pt] &= \frac{(\cosh x)(\cosh x) - (\sinh x)(\sinh x)}{\cosh^2 x} \\[6pt] &= \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} \\[6pt] &= \frac{1}{\cosh^2 x} \\[6pt] &= \text{sech}^2 x \end{align*} \]

\( \textbf{Thm} \)

\( \text{Let } u \text{ be a differentiable function of } x \text{.} \text{} \)

\[ \begin{align*} \int \cosh u\,du &= \sinh u + C \\[6pt] \int \sinh u\,du &= \cosh u + C \\[6pt] \int \text{sech}^2 u\,du &= \tanh u + C \\[6pt] \int \text{csch}^2 u\,du &= -\coth u + C \\[6pt] \int \text{sech } u \tanh u\,du &= -\text{sech } u + C \\[6pt] \int \text{csch } u \coth u\,du &= -\text{csch } u + C \end{align*} \]

Inverse Hyperbolic Functions

\( \textbf{Thm}\)

Function Logarithmic Form Domain
\( \sinh^{-1} x \) \( \ln(x + \sqrt{x^2 + 1}) \) \( (-\infty, \infty) \)
\( \cosh^{-1} x \) \( \ln(x + \sqrt{x^2 - 1}) \) \( [1, \infty) \)
\( \tanh^{-1} x \) \( \frac{1}{2} \ln \left(\frac{1 + x}{1 - x}\right) \) \( (-1, 1) \)
\( \coth^{-1} x \) \( \frac{1}{2} \ln \left(\frac{x + 1}{x - 1}\right) \) \( (-\infty, -1) \cup (1, \infty) \)
\( \sec^{-1} x \) \( \ln \left(\frac{1 + \sqrt{1 - x^2}}{x}\right) \) \( (0, 1] \)
\( \csc^{-1} x \) \( \ln \left(\frac{1 + \sqrt{1 + x^2}}{|x|}\right) \) \( (-\infty, 0) \cup (0, \infty) \)

\( \textbf{Pf } \quad \sinh^{-1} x = \ln(x + \sqrt{x^2 + 1}) \)

\( \text{Let } y = \sinh^{-1} x \text{. Then } x = \sinh y \text{.} \)

\[ x = \frac{e^{y} - e^{-y}}{2} \] \[ 2x = e^{y} - e^{-y} \] \[ 2x e^{y} = e^{2y} - e^{0} \] \[ 2x e^{y} = (e^{y})^2 - 1 \] \[ (e^{y})^2 - 2x e^{y} - 1 = 0 \]

\( \text{Let } z = e^{y} \)

\[ (z)^2 - 2x (z) - 1 = 0 \] \[ z = e^{y} = \frac{-(-2x) \pm \sqrt{(-2x)^2 - 4(1)(-1)}}{2(1)} \] \[ \begin{align*} e^{y} &= \frac{2x \pm \sqrt{4x^2 + 4}}{2} \\[6pt] &= \frac{2x \pm 2\sqrt{x^2 + 1}}{2} \\[6pt] &= x \pm \sqrt{x^2 + 1} \end{align*} \]

\[ e^{y} = x + \sqrt{x^2 + 1} \] \[ y = \ln(x + \sqrt{x^2 + 1}) \] \[ \sinh^{-1} x = \ln(x + \sqrt{x^2 + 1}) \]

\( \displaystyle \textbf{Pf} \quad \tanh^{-1} x = \frac{1}{2} \ln \left(\frac{1 + x}{1 - x}\right) \text{} \)

\( \text{Let } y = \tanh^{-1} x \text{. Then } x = \tanh y \text{, where } -1 < x < 1 \text{.} \)

\[ x = \frac{\sinh y}{\cosh y} = \frac{e^{y} - e^{-y}}{e^{y} + e^{-y}} \] \[ x (e^{y} + e^{-y}) = e^{y} - e^{-y} \] \[ x e^{y} + x e^{-y} = e^{y} - e^{-y} \] \[ x e^{-y} + e^{-y} = e^{y} - x e^{y} \] \[ e^{-y} (x + 1) = e^{y} (1 - x) \] \[ \frac{e^{y}}{e^{-y}} = \frac{x + 1}{1 - x} \] \[ e^{2y} = \frac{1 + x}{1 - x} \] \[ \ln(e^{2y}) = \ln \left(\frac{1 + x}{1 - x}\right) \] \[ 2y = \ln \left(\frac{1 + x}{1 - x}\right) \] \[ y = \frac{1}{2} \ln \left(\frac{1 + x}{1 - x}\right) \] \[ \tanh^{-1} x = \frac{1}{2} \ln \left(\frac{1 + x}{1 - x}\right) \text{} \]

Differentiation and Integration Involving Inverse Hyperbolic Functions

\( \textbf{Thm} \)

\( \text{Let } u \text{ be a differentiable function of } x \text{, and let } a > 0 \text{.} \text{} \)

\( \textbf{A. Differentiation Formulas} \)

\[ \begin{align*} \frac{d}{dx}[\sinh^{-1} u] &= \frac{u'}{\sqrt{u^2 + 1}} \\[6pt] \frac{d}{dx}[\tanh^{-1} u] &= \frac{u'}{1 - u^2} \\[6pt] \frac{d}{dx}[\sec^{-1} u] &= \frac{-u'}{u\sqrt{1 - u^2}} \end{align*} \]
\[ \begin{align*} \frac{d}{dx}[\cosh^{-1} u] &= \frac{u'}{\sqrt{u^2 - 1}} \\[6pt] \frac{d}{dx}[\coth^{-1} u] &= \frac{u'}{1 - u^2} \\[6pt] \frac{d}{dx}[\csc^{-1} u] &= \frac{-u'}{|u|\sqrt{1 + u^2}} \end{align*} \]

\( \textbf{B. Integration Formulas} \)

\[ \begin{align*} \int \frac{du}{\sqrt{u^2 \pm a^2}} &= \ln(u + \sqrt{u^2 \pm a^2}) + C \\[6pt] \int \frac{du}{a^2 - u^2} &= \frac{1}{2a} \ln \left| \frac{a + u}{a - u} \right| + C \quad (\text{for } |u| < a) \\[6pt] \int \frac{du}{u\sqrt{a^2 - u^2}} &= \frac{1}{a} \ln \left| \frac{a + \sqrt{a^2 - u^2}}{u} \right| + C \\[6pt] \int \frac{du}{u\sqrt{a^2 + u^2}} &= -\frac{1}{a} \ln \left| \frac{a + \sqrt{a^2 + u^2}}{u} \right| + C \end{align*} \]

\( \displaystyle \textbf{Pf} \quad \frac{d}{dx}[\sinh^{-1} x] \)

\( \text{Let } y = \sinh^{-1} x \text{. Then } x = \sinh y \text{.} \)

\[ \displaystyle \begin{align*} \frac{dy}{dx} &= \frac{1}{\frac{d}{dy}[\sinh y]} \\[6pt] &= \frac{1}{\cosh y} \end{align*} \]

\( \cosh^2 y - \sinh^2 y = 1 \text{} \text{, we have } \cosh y = \sqrt{1 + \sinh^2 y}\)

\[ \begin{align*} \frac{d}{dx}[\sinh^{-1} x] &= \frac{1}{\sqrt{1 + x^2}} \\[6pt] \Rightarrow \frac{d}{dx}[\sinh^{-1} u] &= \frac{u'}{\sqrt{u^2 + 1}} \end{align*} \]

\( \displaystyle \textbf{Pf} \quad \frac{d}{dx}[\tanh^{-1} x] \)

\( \text{Let } y = \tanh^{-1} x \text{. Then } x = \tanh y \text{.} \)

\[ \begin{align*} \frac{dy}{dx} &= \frac{1}{\frac{d}{dy}[\tanh y]} \\[6pt] &= \frac{1}{\text{sech}^2 y} \end{align*} \]

\( \tanh^2 y + \text{sech}^2 y = 1 \text{} \text{, we have } \text{sech}^2 y = 1 - \tanh^2 y \text{.} \)

\[ \begin{align*} \frac{d}{dx}[\tanh^{-1} x] &= \frac{1}{1 - x^2} \\[6pt] \frac{d}{dx}[\tanh^{-1} u] &= \frac{u'}{1 - u^2} \end{align*} \]

\( \displaystyle \textbf{EX} \quad \frac{d}{dx}\sinh^{-1}(\tan x) \)

\( \text{Let } u = \tan x ,\quad u' = \sec^2 x \)

\[ \displaystyle \begin{align*} &\frac{d}{dx}[\sinh^{-1}(\tan x)] \\[6pt] =& \frac{d}{dx}[\sinh^{-1}(u)] \\[6pt] =& \frac{\sec^2 x}{\sqrt{(\tan x)^2 + 1}} \\[6pt] =& \frac{\sec^2 x}{\sqrt{\sec^2 x}} \\[6pt] =& \frac{\sec^2 x}{|\sec x|} = |\sec x| \end{align*} \]

\( \displaystyle \textbf{EX} \quad \frac{d}{dx}\tanh^{-1}(\sin x) \)

\[ \displaystyle \begin{align*} & \frac{d}{dx}\tanh^{-1}(\sin x) \\[6pt] =& \frac{1}{1 - \sin^{2}x}\frac{d}{dx}\sin x \\[6pt] =& \frac{\cos x}{\cos^{2}x} \\[6pt] =& \sec x \\[6pt] \Leftrightarrow \int \sec x \, & dx = \tanh^{-1}(\sin x) + C \end{align*} \]

8.1 Basic Intergration Rules

Review of Basic Integration Rules (a > 0)

\begin{align*} 1. & \int k f(u) \, du = k \int f(u) \, du \\[6pt] 2. & \int [f(u) \pm g(u)] \, du = \int f(u) \, du \pm \int g(u) \, du \\[6pt] 3. & \int 1 \, du = u + C \\[6pt] 4. & \int u^n \, du = \frac{u^{n+1}}{n+1} + C, \quad n \neq -1 \\[6pt] 5. & \int \frac{du}{u} = \ln|u| + C \\[6pt] 6. & \int e^u \, du = e^u + C \\[6pt] 7. & \int a^u \, du = \left( \frac{1}{\ln a} \right) a^u + C \\[6pt] 8. & \int \sin u \, du = -\cos u + C \\[6pt] 9. & \int \cos u \, du = \sin u + C \\[6pt] 10. & \int \tan u \, du = -\ln|\cos u| + C \\[6pt] 11. & \int \cot u \, du = \ln|\sin u| + C \\[6pt] 12. & \int \sec u \, du = \ln|\sec u + \tan u| + C \\[6pt] 13. & \int \csc u \, du = -\ln|\csc u + \cot u| + C \\[6pt] 14. & \int \sec^2 u \, du = \tan u + C \\[6pt] 15. & \int \csc^2 u \, du = -\cot u + C \\[6pt] 16. & \int \sec u \tan u \, du = \sec u + C \\[6pt] 17. & \int \csc u \cot u \, du = -\csc u + C \\[6pt] 18. & \int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1} \frac{u}{a} + C \\[6pt] 19. & \int \frac{du}{a^2 + u^2} = \frac{1}{a} \tan^{-1} \frac{u}{a} + C \\[6pt] 20. & \int \frac{du}{u\sqrt{u^2 - a^2}} = \frac{1}{a} \sec^{-1} \frac{|u|}{a} + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int_0^1 \frac{x+3}{\sqrt{4-x^2}} \, dx \)

\[ \int_0^1 \frac{x+3}{\sqrt{4-x^2}} \, dx = \int_0^1 \frac{x}{\sqrt{4-x^2}} \, dx + \int_0^1 \frac{3}{\sqrt{4-x^2}} \, dx \]
\[ \text{Let } u = 4-x^2 \implies du = -2x \, dx \implies x \, dx = -\frac{1}{2} du \]
\[ \begin{align*} &\int \frac{3}{\sqrt{2^2-x^2}} \, dx + \int \frac{x}{\sqrt{4-x^2}} \, dx \\[6pt] &= 3 \sin^{-1} \frac{x}{2} -\frac{1}{2} \int u^{-1/2} \, du \\[6pt] &=3 \sin^{-1} \frac{x}{2} -\frac{1}{2} (2u^{1/2}) \\[6pt] &=3 \sin^{-1} \frac{x}{2} -\sqrt{4-x^2} \\[6pt] &= \Rightarrow \left[ -\sqrt{4-x^2} + 3 \sin^{-1} \frac{x}{2} \right]_0^1 \\[6pt] &= \left( -\sqrt{4-1^2} + 3 \sin^{-1} \frac{1}{2} \right) - \left( -\sqrt{4-0^2} + 3 \sin^{-1} 0 \right) \\[6pt] &= \left( -\sqrt{3} + 3 \cdot \frac{\pi}{6} \right) - ( -2 + 0 ) \\[6pt] &= 2 - \sqrt{3} + \frac{\pi}{2} \end{align*} \]

8.2 Intergration by Parts

\( \displaystyle \textbf{Thm} \)

\( \text{If } u \text{ and } v \text{ are functions of } x \text{ and have continious derivatives , then}\)

\[ \int u \, dv = uv - \int v \, du \]

\( \displaystyle \textbf{Rule} \quad \text{LIATE Rule for Choosing } u \)

依照下列順序，排在愈前方的函數愈優先設為 \( u \)

\begin{align*} \textbf{L} & \quad \text{Logarithmic functions} \quad (\ln x, \log_a x) \\[6pt] \textbf{I} & \quad \text{Inverse trigonometric functions} \quad (\arcsin x) \\[6pt] \textbf{A} & \quad \text{Algebraic functions} \quad (x^n, 2x, \text{polynomials}) \\[6pt] \textbf{T} & \quad \text{Trigonometric functions} \quad (\sin x, \cos x) \\[6pt] \textbf{E} & \quad \text{Exponential functions} \quad (e^x, a^x) \end{align*}

\( \displaystyle \textbf{EX} \quad \int x \, \sin \, x \, dx \)

\( \text{Let } u = x ,\quad dv = \sin x \, dx du = 1 \, dx ,\quad v = - \cos x \)

\[ \begin{align*} & \int x \, \sin \, x \, dx \\[6pt] &= -x\cos x + \int \cos x \, dx \\[6pt] &= -x \cos x + \sin x + C \end{align*} \]

\( \displaystyle \textbf{EX} \quad \int \ln x \, dx \)

\begin{align*} \text{Let} \quad u &= \ln x \quad \Rightarrow \quad du = \frac{1}{x} \, dx \\ dv &= dx \quad \Rightarrow \quad v = x \end{align*}
\begin{align*} \Rightarrow \int \ln x \, dx &= (\ln x)(x) - \int x \cdot \left( \frac{1}{x} \right) \, dx \\ &= x \ln x - \int 1 \, dx \\ &= x \ln x - x + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int x^2 \ln(x+1) \, dx \)

\begin{align*} \text{Let} \quad u &= \ln(x+1) \quad \Rightarrow \quad du = \frac{1}{x+1} \, dx \\[6pt] dv &= x^2 \, dx \quad \Rightarrow \quad v = \frac{1}{3}x^3 \end{align*}
\begin{align*} &\Rightarrow \int x^2 \ln(x+1) \, dx \\[6pt] &= \frac{1}{3}x^3 \ln(x+1) - \int \frac{1}{3}x^3 \cdot \frac{1}{x+1} , dx \\[6pt] &= \frac{1}{3}x^3 \ln(x+1) - \frac{1}{3} \int \frac{x^3}{x+1} , dx \\[6pt] &= \frac{1}{3}x^3 \ln(x+1) - \frac{1}{3} \int \left( x^2 - x + 1 - \frac{1}{x+1} \right) , dx \\[6pt] &= \frac{1}{3}x^3 \ln(x+1) - \frac{1}{3} \left( \frac{1}{3}x^3 - \frac{1}{2}x^2 + x - \ln|x+1| \right) + C \\[6pt] &= \frac{1}{3}x^3 \ln(x+1) - \frac{1}{9}x^3 + \frac{1}{6}x^2 - \frac{1}{3}x + \frac{1}{3}\ln|x+1| + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int e^x \sin x \, dx \)

\begin{align*} \text{Let} \quad u &= \sin x \quad \Rightarrow \quad du = \cos x \, dx \\[6pt] dv &= e^x , dx \quad \Rightarrow \quad v = e^x \end{align*}
\begin{align*} &\Rightarrow \int e^x \sin x \, dx \\[6pt] &= e^x \sin x - \int e^x \cos x \, dx \end{align*}
\begin{align*} \text{Let} \quad u &= \cos x \quad \Rightarrow \quad du = -\sin x \, dx \\[6pt] dv &= e^x \, dx \quad \Rightarrow \quad v = e^x \end{align*}
\begin{align*} &\Rightarrow \int e^x \sin x \, dx \\[6pt] &= e^x \sin x - \left( e^x \cos x - \int e^x (-\sin x) , dx \right) \\[6pt] &= e^x \sin x - e^x \cos x - \int e^x \sin x \, dx \\[6pt] &\Rightarrow 2 \int e^x \sin x \, dx = e^x (\sin x - \cos x) \\[6pt] &= \frac{e^x (\sin x - \cos x)}{2} + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int x^2 e^x \, dx \)

\begin{align*} \text{Let} \quad u &= x^2 \quad \Rightarrow \quad du = 2x \, dx \\[6pt] dv &= e^x \, dx \quad \Rightarrow \quad v = e^x \end{align*}
\begin{align*} &\Rightarrow \int x^2 e^x \, dx \\[6pt] &= x^2 e^x - \int 2x e^x \, dx \\[6pt] &= x^2 e^x - 2 \int x e^x \, dx \end{align*}
\begin{align*} \text{Let} \quad u &= x \quad \Rightarrow \quad du = dx \\[6pt] dv &= e^x \, dx \quad \Rightarrow \quad v = e^x \end{align*}
\begin{align*} &\Rightarrow x^2 e^x - 2 \left( x e^x - \int e^x \, dx \right) \\[6pt] &= x^2 e^x - 2 (x e^x - e^x) + C \\[6pt] &= x^2 e^x - 2xe^x + 2e^x + C \\[6pt] &= e^x (x^2 - 2x + 2) + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int x^3 \cos(2x) \, dx \)

使用表格法 (Tabular Integration)

\begin{align*} \begin{array}{c|c|c} \text{Sign} & u \text{ (Diff)} & dv \text{ (Int)} \\ \hline + & x^3 & \cos(2x) \\ - & 3x^2 & \frac{1}{2}\sin(2x) \\ + & 6x & -\frac{1}{4}\cos(2x) \\ - & 6 & -\frac{1}{8}\sin(2x) \\ + & 0 & \frac{1}{16}\cos(2x) \end{array} \end{align*}
\begin{align*} &\Rightarrow \int x^3 \cos(2x) \, dx \\[6pt] &= (x^3)\left( \frac{1}{2}\sin(2x) \right) - (3x^2)\left( -\frac{1}{4}\cos(2x) \right) + (6x)\left( -\frac{1}{8}\sin(2x) \right) - (6)\left( \frac{1}{16}\cos(2x) \right) + C \\[6pt] &= \frac{1}{2}x^3 \sin(2x) + \frac{3}{4}x^2 \cos(2x) - \frac{3}{4}x \sin(2x) - \frac{3}{8}\cos(2x) + C \\[6pt] &= \left( \frac{1}{2}x^3 - \frac{3}{4}x \right) \sin(2x) + \left( \frac{3}{4}x^2 - \frac{3}{8} \right) \cos(2x) + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \arctan x \, dx \)

\begin{align*} \text{Let} \quad u &= \arctan x \quad \Rightarrow \quad du = \frac{1}{1+x^2} \, dx \\[6pt] dv &= dx \quad \Rightarrow \quad v = x \end{align*}
\begin{align*} &\Rightarrow \int \arctan x \, dx \\[6pt] &= x \arctan x - \int \frac{x}{1+x^2} \, dx \\[6pt] &= x \arctan x - \frac{1}{2} \int \frac{2x}{1+x^2} \, dx \\[6pt] &= x \arctan x - \frac{1}{2} \ln(1+x^2) + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \cos(\sqrt{x}) \, dx \)

\begin{align*} \text{Let} \quad w &= \sqrt{x} = x^{1/2} \quad \Rightarrow \quad dw = \frac{1}{2\sqrt{x}} \, dx \\[6pt] dx &= 2\sqrt{x} \, dw = 2w \, dw \end{align*}
\begin{align*} &\Rightarrow \int \cos(\sqrt{x}) \, dx \\[6pt] &= \int \cos(w) \cdot 2w \, dw \\[6pt] &= 2 \int w \cos w \, dw \end{align*}
\begin{align*} \text{Let} \quad u &= w \quad \Rightarrow \quad du = dw \\[6pt] dv &= \cos w \, dw \quad \Rightarrow \quad v = \sin w \end{align*}
\begin{align*} &\Rightarrow 2 \left( w \sin w - \int \sin w \, dw \right) \\[6pt] &= 2 (w \sin w + \cos w) + C \\[6pt] &= 2\sqrt{x} \sin(\sqrt{x}) + 2\cos(\sqrt{x}) + C \end{align*}

8.3 Trigonometric Integrals

Integrals Involving Powers of Sine and Cosine

1. 當 \(\sin\) 的次方為奇數時： 拆出一個 \(\sin x\)，將其餘轉為 \(\cos\)

\begin{align*} \int \sin^{2k+1} x \cos^n x \, dx &= \int (\sin^2 x)^k \cos^n x \sin x \, dx \\[6pt] &= \int (1 - \cos^2 x)^k \cos^n x \sin x \, dx \end{align*}

2. 當 \(\cos\) 的次方為奇數時： 拆出一個 \(\cos x\)，將其餘轉為 \(\sin\)

\begin{align*} \int \sin^m x \cos^{2k+1} x \, dx &= \int \sin^m x (\cos^2 x)^k \cos x \, dx \\[6pt] &= \int \sin^m x (1 - \sin^2 x)^k \cos x \, dx \end{align*}

3. 當 \(\sin\) 與 \(\cos\) 均為偶數時： 使用降次公式

\begin{align*} \sin^2 x &= \frac{1 - \cos 2x}{2} \\[6pt] \cos^2 x &= \frac{1 + \cos 2x}{2} \end{align*}

\( \displaystyle \textbf{EX} \quad \int \sin^2 x \cos x \, dx \)

\begin{align*} \text{Let} \quad u &= \sin x \quad \Rightarrow \quad du = \cos x \, dx \end{align*}
\begin{align*} & \int (\sin^2 x) (\cos x \, dx) \\[6pt] =& \int u^2 \, du \\[6pt] =& \frac{1}{3}u^3 + C \\[6pt] =& \frac{1}{3}\sin^3 x + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \sin^2 x \cos^2 x \, dx \)

\begin{align*} \sin^2 x &= \frac{1 - \cos 2x}{2} \\[6pt] \cos^2 x &= \frac{1 + \cos 2x}{2} \end{align*}
\begin{align*} &\Rightarrow \int \left( \frac{1 - \cos 2x}{2} \right) \left( \frac{1 + \cos 2x}{2} \right) \, dx \\[6pt] &= \frac{1}{4} \int (1 - \cos^2 2x) \, dx \\[6pt] &= \frac{1}{4} \int \sin^2 2x \, dx \end{align*}
\begin{align*} \sin^2 2x = \frac{1 - \cos 4x}{2} \end{align*}
\begin{align*} &\Rightarrow \frac{1}{4} \int \frac{1 - \cos 4x}{2} \, dx \\[6pt] &= \frac{1}{8} \int (1 - \cos 4x) \, dx \\[6pt] &= \frac{1}{8} \left( x - \frac{1}{4} \sin 4x \right) + C \\[6pt] &= \frac{1}{8}x - \frac{1}{32}\sin 4x + C \end{align*}

Integrals Involving Powers of Secant and Tangent

1. 當 \(\sec\) 的次方為偶數時： 拆出 \(\sec^2 x\)，其餘轉為 \(\tan\)

\begin{align*} \int \sec^{2k} x \tan^n x \, dx &= \int (\sec^2 x)^{k-1} \tan^n x \sec^2 x \, dx \\[6pt] &= \int (1 + \tan^2 x)^{k-1} \tan^n x \sec^2 x \, dx \end{align*}

2. 當 \(\tan\) 的次方為奇數時： 拆出 \(\sec x \tan x\)，其餘轉為 \(\sec\)

\begin{align*} \int \sec^m x \tan^{2k+1} x \, dx &= \int \sec^{m-1} x (\tan^2 x)^k \sec x \tan x \, dx \\[6pt] &= \int \sec^{m-1} x (\sec^2 x - 1)^k \sec x \tan x \, dx \end{align*}

3. 只有 \(\tan\) 且為偶數次方時：

\begin{align*} \int \tan^n x \, dx &= \int \tan^{n-2} x (\tan^2 x) \, dx \\[6pt] &= \int \tan^{n-2} x (\sec^2 x - 1) \, dx \end{align*}

4. 當 \(\sec\) 為奇數次方時： 使用分部積分

\begin{align*} \int \sec^m x \, dx \quad (m \text{ is odd}) \end{align*}

\( \displaystyle \textbf{EX} \quad \int \tan^3 x \sec^4 x \, dx \)

\begin{align*} \text{Let} \quad u &= \sec x \quad \Rightarrow \quad du = \sec x \tan x \, dx \\[6pt] \tan^2 x &= \sec^2 x - 1 = u^2 - 1 \end{align*}
\begin{align*} &\Rightarrow \int (\tan^2 x) \cdot (\sec^3 x) \cdot (\sec x \tan x \, dx) \\[6pt] &= \int (u^2 - 1) \cdot u^3 \, du \\[6pt] &= \int (u^5 - u^3) \, du \\[6pt] &= \frac{1}{6}u^6 - \frac{1}{4}u^4 + C \\[6pt] &= \frac{1}{6}\sec^6 x - \frac{1}{4}\sec^4 x + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \tan^{1/3} x \sec^4 x \, dx \)

\begin{align*} \text{Let} \quad u &= \tan x \quad \Rightarrow \quad du = \sec^2 x \, dx \\[6pt] \sec^2 x &= 1 + \tan^2 x = 1 + u^2 \end{align*}
\begin{align*} &\Rightarrow \int \tan^{1/3} x \cdot \sec^2 x \cdot (\sec^2 x \, dx) \\[6pt] &= \int u^{1/3} (1 + u^2) \, du \\[6pt] &= \int (u^{1/3} + u^{7/3}) \, du \\[6pt] &= \frac{u^{4/3}}{4/3} + \frac{u^{10/3}}{10/3} + C \\[6pt] &= \frac{3}{4} u^{4/3} + \frac{3}{10} u^{10/3} + C \\[6pt] &= \frac{3}{4} \tan^{4/3} x + \frac{3}{10} \tan^{10/3} x + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \sec^3 x \, dx \)

\begin{align*} \text{Let} \quad u &= \sec x \quad \Rightarrow \quad du = \sec x \tan x \, dx \\[6pt] dv &= \sec^2 x \, dx \quad \Rightarrow \quad v = \tan x \end{align*}
\begin{align*} \Rightarrow \int \sec^3 x \, dx &= \sec x \tan x - \int \tan x (\sec x \tan x) \, dx \\[6pt] &= \sec x \tan x - \int \sec x \tan^2 x \, dx \\[6pt] &= \sec x \tan x - \int \sec x (\sec^2 x - 1) \, dx \\[6pt] &= \sec x \tan x - \left( \int \sec^3 x \, dx - \int \sec x \, dx \right) \end{align*}
\begin{align*} \Rightarrow 2 \int \sec^3 x \, dx &= \sec x \tan x + \int \sec x \, dx \\[6pt] &= \sec x \tan x + \ln|\sec x + \tan x| + C \\[6pt] \Rightarrow \int \sec^3 x \, dx &= \frac{1}{2} \left( \sec x \tan x + \ln|\sec x + \tan x| \right) + C\end{align*}

\( \displaystyle \textbf{EX} \quad \int \tan^3 x \, dx \quad \int \tan^4 x \, dx \quad \int \tan^5 x \, dx \)

\[ \displaystyle \text{Let} \quad u = \tan x, \, du = \sec^2 x \, dx \]
\begin{align*} \int \tan^3 x \, dx &= \int \tan x (\tan^2 x) \, dx \\[6pt] &= \int \tan x (\sec^2 x - 1) \, dx \\[6pt] &= \int \tan x \sec^2 x \, dx - \int \tan x \, dx \end{align*} \begin{align*} &\Rightarrow \int u \, du - \ln|\sec x| \\[6pt] &= \frac{1}{2}\tan^2 x - \ln|\sec x| + C \end{align*}
\begin{align*} \int \tan^4 x \, dx &= \int \tan^2 x (\sec^2 x - 1) \, dx \\[6pt] &= \int \tan^2 x \sec^2 x \, dx - \int \tan^2 x \, dx \\[6pt] &= \int \tan^2 x \sec^2 x \, dx - \int (\sec^2 x - 1) \, dx \end{align*} \begin{align*} &= \int u^2 \, du - (\tan x - x) \\[6pt] &= \frac{1}{3}\tan^3 x - \tan x + x + C \end{align*}
\begin{align*} \int \tan^5 x \, dx &= \int \tan^3 x (\sec^2 x - 1) \, dx \\[6pt] &= \int \tan^3 x \sec^2 x \, dx - \int \tan^3 x \, dx \\[6pt] &= \frac{1}{4}\tan^4 x - \left( \frac{1}{2}\tan^2 x - \ln|\sec x| \right) + C \\[6pt] &= \frac{1}{4}\tan^4 x - \frac{1}{2}\tan^2 x + \ln|\sec x| + C \end{align*}

Wallis's Formulas

對於定積分 \( \displaystyle \int_0^{\pi/2} \cos^n x \, dx \text{ 和 } \int_0^{\pi/2} \sin^n x \, dx \)：

\[ \begin{align*} \text{If } n \text{ is odd: } & \quad \left( \frac{2}{3} \right) \left( \frac{4}{5} \right) \left( \frac{6}{7} \right) \cdots \left( \frac{n-1}{n} \right) \\[6pt] \text{If } n \text{ is even: } & \quad \left( \frac{1}{2} \right) \left( \frac{3}{4} \right) \left( \frac{5}{6} \right) \cdots \left( \frac{n-1}{n} \right) \left( \frac{\pi}{2} \right) \end{align*} \]

\( \displaystyle \textbf{EX} \quad \int_0^{\pi/2} \sin^5 x \, dx \)

\begin{align*} \int_0^{\pi/2} \sin^5 x \, dx &= \left( \frac{n-1}{n} \right) \left( \frac{n-3}{n-2} \right) \cdots \\[6pt] &= \left( \frac{4}{5} \right) \left( \frac{2}{3} \right) \\[6pt] &= \frac{8}{15} \end{align*}

Product-to-Sum Formulas with sines and cosines 和差化積公式

\begin{align*} \sin mx \sin nx &= \frac{1}{2} \left[ \cos(m-n)x - \cos(m+n)x \right] \\[6pt] \sin mx \cos nx &= \frac{1}{2} \left[ \sin(m-n)x + \sin(m+n)x \right] \\[6pt] \cos mx \cos nx &= \frac{1}{2} \left[ \cos(m-n)x + \cos(m+n)x \right] \end{align*}

\( \displaystyle \textbf{EX} \quad \int \sin 5x \cos 4x \, dx \)

\begin{align*} &\int \sin 5x \cos 4x \, dx \\[6pt] =& \frac{1}{2} \left( \int \sin x \, dx + \int \sin 9x \, dx \right) \\[6pt] =& \frac{1}{2} \left( -\cos x - \frac{\cos 9x}{9} \right) + C \\[6pt] =& -\frac{\cos x}{2} - \frac{\cos 9x}{18} + C \end{align*}

8.4 Trigonometric Substitution

Trigonometric Substitution (a > 0)

\( \sqrt{a^2 - u^2} \quad \text{, Let } u = a \sin \theta \)

\begin{align*} \sqrt{a^2 - u^2} &= \sqrt{a^2 - a^2\sin^2 \theta} = a \cos \theta \\[6pt] du &= a \cos \theta \, d\theta \end{align*}

\( \sqrt{a^2 + u^2} \quad \text{, Let } u = a \tan \theta \)

\begin{align*} \sqrt{a^2 + u^2} &= \sqrt{a^2 + a^2\tan^2 \theta} = a \sec \theta \\[6pt] du &= a \sec^2 \theta \, d\theta \end{align*}

\( \sqrt{u^2 - a^2} \quad \text{, Let } u = a \sec \theta \)

\begin{align*} \sqrt{u^2 - a^2} &= \sqrt{a^2\sec^2 \theta - a^2} = a \tan \theta \\[6pt] du &= a \sec \theta \tan \theta \, d\theta \end{align*}

\( \displaystyle \textbf{Proof} \quad \text{Area of an Ellipse } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

\begin{align*} \text{Let} \quad x &= a \sin \theta \quad \Rightarrow \quad dx = a \cos \theta \, d\theta \\[6pt] \sqrt{a^2 - x^2} &= \sqrt{a^2 - a^2\sin^2 \theta} = a \cos \theta \end{align*} \begin{align*} \text{When } x = 0: & \quad 0 = a \sin \theta \quad \Rightarrow \quad \theta = 0 \\[6pt] \text{When } x = a: & \quad a = a \sin \theta \quad \Rightarrow \quad \theta = \pi/2 \end{align*}
\begin{align*} A_{\text{quadrant}} &= \frac{b}{a} \int_0^a \sqrt{a^2 - x^2} \, dx \\[6pt] &= \frac{b}{a} \int_0^{\pi/2} (a \cos \theta) (a \cos \theta \, d\theta) \\[6pt] &= ab \int_0^{\pi/2} \cos^2 \theta \, d\theta \end{align*}
\begin{align*} \int_0^{\pi/2} \cos^2 \theta \, d\theta &= \int_0^{\pi/2} \frac{1 + \cos 2\theta}{2} \, d\theta \\[6pt] &= \left[ \frac{1}{2}\theta + \frac{1}{4}\sin 2\theta \right]_0^{\pi/2} \\[6pt] &= \left( \frac{\pi}{4} + 0 \right) - (0 + 0) = \frac{\pi}{4} \end{align*}
\begin{align*} A_{\text{ellipse}} &= 4 \times A_{\text{quadrant}} \\[6pt] &= 4 \times \left( ab \cdot \frac{\pi}{4} \right) \\[6pt] &= \pi ab \end{align*}

\( \displaystyle \textbf{EX} \quad \int \frac{x^2}{\sqrt{4-x^2}} \, dx \)

\begin{align*} \text{Let} \quad x &= 2 \sin \theta \quad \Rightarrow \quad dx = 2 \cos \theta \, d\theta \\[6pt] \sqrt{4-x^2} &= \sqrt{4-4\sin^2 \theta} = 2 \cos \theta \end{align*}
\begin{align*} &\Rightarrow \int \frac{(2\sin \theta)^2}{2\cos \theta} \cdot (2\cos \theta \, d\theta) \\[6pt] &= \int 4\sin^2 \theta \, d\theta \\[6pt] &= 4 \int \frac{1 - \cos 2\theta}{2} \, d\theta \\[6pt] &= 2 \int (1 - \cos 2\theta) \, d\theta \\[6pt] &= 2\theta - \sin 2\theta + C \\[6pt] &= 2\theta - 2\sin \theta \cos \theta + C \end{align*}
\begin{align*} \sin \theta &= \frac{x}{2} \quad \Rightarrow \quad \theta = \arcsin\left(\frac{x}{2}\right) \\[6pt] \cos \theta &= \frac{\sqrt{4-x^2}}{2} \end{align*}
\begin{align*} &\Rightarrow 2 \arcsin\left(\frac{x}{2}\right) - 2 \left( \frac{x}{2} \right) \left( \frac{\sqrt{4-x^2}}{2} \right) + C \\[6pt] &= 2 \arcsin\left(\frac{x}{2}\right) - \frac{x\sqrt{4-x^2}}{2} + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \sqrt{9-x^2} \, dx \)

\begin{align*} \text{Let} \quad x &= 3 \sin \theta \quad \Rightarrow \quad dx = 3 \cos \theta \, d\theta \\[6pt] \sqrt{9-x^2} &= 3 \cos \theta \end{align*}
\begin{align*} &\Rightarrow \int (3 \cos \theta) (3 \cos \theta) \, d\theta = 9 \int \cos^2 \theta \, d\theta \\[6pt] &= 9 \int \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{9}{2} \left( \theta + \frac{1}{2}\sin 2\theta \right) + C \\[6pt] &= \frac{9}{2} \theta + \frac{9}{2} \sin \theta \cos \theta + C \\[6pt] &= \frac{9}{2} \arcsin\left(\frac{x}{3}\right) + \frac{x\sqrt{9-x^2}}{2} + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \frac{1}{\sqrt{x^2+4}} \, dx \)

\begin{align*} \text{Let} \quad x &= 2 \tan \theta \quad \Rightarrow \quad dx = 2 \sec^2 \theta \, d\theta \\[6pt] \sqrt{x^2+4} &= 2 \sec \theta \end{align*}
\begin{align*} &\Rightarrow \int \frac{2 \sec^2 \theta}{2 \sec \theta} \, d\theta = \int \sec \theta \, d\theta \\[6pt] &= \ln|\sec \theta + \tan \theta| + C \\[6pt] &= \ln\left| \frac{\sqrt{x^2+4}}{2} + \frac{x}{2} \right| + C \\[6pt] &= \ln|x + \sqrt{x^2+4}| + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \frac{x}{\sqrt{3-2x-x^2}} \, dx \)

\begin{align*} &\int \frac{x}{\sqrt{3-2x-x^2}} \, dx \\[6pt] =& \int \frac{x+1}{\sqrt{4-(x+1)^2}} \, dx - \int \frac{1}{\sqrt{4-(x+1)^2}} \, dx \\[6pt] \end{align*}
\[ \begin{align*} \text{Let} \quad u &= 3-2x-x^2, \\[6pt] du &= (-2-2x)dx = -2(x+1)dx \end{align*} \]
\begin{align*} &\int \frac{x+1}{\sqrt{4-(x+1)^2}} \, dx \\[6pt] =& \int \frac{x+1}{\sqrt{u}} \cdot \frac{du}{-2(x+1)} \\[6pt] =& -\frac{1}{2} \int u^{-1/2} \, du \\[6pt] =& -\frac{1}{2} (2u^{1/2}) \\[6pt] =& -\sqrt{3-2x-x^2} \end{align*}
\[ \text{Let } x+1 = 2\sin\theta, \quad dx = 2\cos\theta \, d\theta \]
\begin{align*} &\int \frac{1}{\sqrt{4-(x+1)^2}} \, dx \\[6pt] =&\int \frac{2\cos\theta}{2\cos\theta} \, d\theta \\[6pt] =& \theta = \arcsin\left(\frac{x+1}{2}\right) \end{align*}
\[ \begin{align*} \Rightarrow&\int \frac{x}{\sqrt{3-2x-x^2}} \, dx \\[6pt] =& -\sqrt{3-2x-x^2} - \arcsin\left(\frac{x+1}{2}\right) + C \end{align*} \]

\( \displaystyle \textbf{EX} \quad \int \frac{\sqrt{x^2-3}}{x} \, dx \)

\begin{align*} \text{Let} \quad x &= \sqrt{3} \sec \theta \\[6pt] \Rightarrow \quad dx &= \sqrt{3} \sec \theta \tan \theta \, d\theta \\[6pt] \sqrt{x^2-3} &= \sqrt{3} \tan \theta \end{align*}
\begin{align*} &\Rightarrow \int \frac{\sqrt{3} \tan \theta}{\sqrt{3} \sec \theta} \cdot (\sqrt{3} \sec \theta \tan \theta) \, d\theta \\[6pt] &= \sqrt{3} \int \tan^2 \theta \, d\theta = \sqrt{3} \int (\sec^2 \theta - 1) \, d\theta \\[6pt] &= \sqrt{3} (\tan \theta - \theta) + C \\[6pt] &= \sqrt{x^2-3} - \sqrt{3} \text{arcsec}\left(\frac{x}{\sqrt{3}}\right) + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int_1^2 x^3 \sqrt{x^2-1} \, dx \)

\begin{align*} \text{Let} \quad x &= \sec \theta \\[6pt] \Rightarrow \quad dx &= \sec \theta \tan \theta \, d\theta \\[6pt] \sqrt{x^2-1} &= \tan \theta \\[6pt] & x = 1 \Rightarrow \sec \theta = 1 \Rightarrow \theta = 0 \\[6pt] & x = 2 \Rightarrow \sec \theta = 2 \Rightarrow \theta = \frac{\pi}{3} \end{align*}
\begin{align*} &\Rightarrow \int_0^{\pi/3} (\sec^3 \theta) (\tan \theta) (\sec \theta \tan \theta \, d\theta) \\[6pt] &= \int_0^{\pi/3} \sec^4 \theta \tan^2 \theta \, d\theta \\[6pt] &= \int_0^{\pi/3} \sec^2 \theta \tan^2 \theta (\sec^2 \theta \, d\theta) \\[6pt] &= \int_0^{\pi/3} (1 + \tan^2 \theta) \tan^2 \theta (\sec^2 \theta \, d\theta) \end{align*}
\begin{align*} u &= \tan \theta \quad \Rightarrow \quad du = \sec^2 \theta \, d\theta \\[6pt] &\theta = 0 \Rightarrow u = 0 \\[6pt] &\theta = \frac{\pi}{3} \Rightarrow u = \sqrt{3} \end{align*}
\begin{align*} &\Rightarrow \int_0^{\sqrt{3}} (1 + u^2) u^2 \, du \\[6pt] &= \int_0^{\sqrt{3}} (u^2 + u^4) \, du \\[6pt] &= \left[ \frac{1}{3}u^3 + \frac{1}{5}u^5 \right]_0^{\sqrt{3}} \\[6pt] &= \frac{1}{3}(\sqrt{3})^3 + \frac{1}{5}(\sqrt{3})^5 \\[6pt] &= \frac{1}{3}(3\sqrt{3}) + \frac{1}{5}(9\sqrt{3}) \\[6pt] &= \sqrt{3} + \frac{9\sqrt{3}}{5} = \frac{14\sqrt{3}}{5} \end{align*}

8.5 Partial Fraction

Partial Fraction Decomposition Cases

相異線性因式 分母分解後為單純的一次方因式相乘

\begin{align*} \frac{5x+3}{(x-1)(x+2)} &= \frac{A}{x-1} + \frac{B}{x+2} \end{align*}

重複線性因式 某個一次方項出現了多次（如平方、立方）

\begin{align*} \frac{x^2+1}{(x-1)^3} &= \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} \end{align*}

不可約二次因式 分母含有無法再分解成實數一次方的二次項

\begin{align*} \frac{2x+1}{(x-1)(x^2+4)} &= \frac{A}{x-1} + \frac{Bx+C}{x^2+4} \end{align*}

Ps: 二次項上面的分子設為 \( Bx+C \)

重複不可約二次因式

\begin{align*} \frac{1}{x(x^2+1)^2} &= \frac{A}{x} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2} \end{align*}

\( \displaystyle \textbf{EX} \quad \int \frac{1}{(x-2)(x+2)} \, dx \)

\[ \displaystyle \frac{1}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2} \]
\begin{align*} 1 &= A(x+2) + B(x-2) \\[6pt] \text{Let } x=2 &\Rightarrow 1 = 4A \Rightarrow A = 1/4 \\[6pt] \text{Let } x=-2 &\Rightarrow 1 = -4B \Rightarrow B = -1/4 \end{align*}
\begin{align*} &\Rightarrow \int \left( \frac{1/4}{x-2} - \frac{1/4}{x+2} \right) dx \\[6pt] &= \frac{1}{4} \ln|x-2| - \frac{1}{4} \ln|x+2| + C \\[6pt] &= \frac{1}{4} \ln \left| \frac{x-2}{x+2} \right| + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \frac{(x^3-2x^2) - 4}{x^3-2x^2} \, dx \)

\begin{align*} &\int \frac{x^3-2x^2-4}{x^3-2x^2} \, dx \\[6pt] =& \int 1 \, dx - \int \frac{4}{x^2(x-2)} \, dx \\[6pt] =& \int 1 \, dx - \int \left( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-2} \, dx \right) \\[6pt] \end{align*}
\[ 4 = Ax(x-2) + B(x-2) + Cx^2 \]\[ \Rightarrow B=-2, C=1, A=-1 \]
\begin{align*} &\Rightarrow \int \left[ 1 - \left( \frac{-1}{x} + \frac{-2}{x^2} + \frac{1}{x-2} \right) \right] dx \\[6pt] &= \int \left( 1 + \frac{1}{x} + \frac{2}{x^2} - \frac{1}{x-2} \right) dx \\[6pt] &= x + \ln|x| - \frac{2}{x} - \ln|x-2| + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \frac{1}{(x+1)(x^2+1)} \, dx \)

\[ \frac{1}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1} \]
\begin{align*} 1 &= A(x^2+1) + (Bx+C)(x+1) \\[6pt] \text{Let } x=-1 &\Rightarrow 1 = 2A \Rightarrow A = 1/2 \\[6pt] \Rightarrow A+B=0 &\Rightarrow B = -1/2 \\[6pt] \Rightarrow A+C=1 &\Rightarrow C = 1/2 \end{align*}
\begin{align*} &\Rightarrow \int \left( \frac{1/2}{x+1} + \frac{-1/2x + 1/2}{x^2+1} \right) dx \\[6pt] &= \frac{1}{2}\ln|x+1| - \frac{1}{4}\int \frac{2x}{x^2+1}dx + \frac{1}{2}\int \frac{1}{x^2+1}dx \\[6pt] &= \frac{1}{2}\ln|x+1| - \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\arctan(x) + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \frac{1}{\sqrt{1+e^x}} \, dx \)

\[ \begin{align} u &= \sqrt{1+e^x} \Rightarrow u^2 = 1+e^x \Rightarrow e^x = u^2-1 \\[6pt] e^x dx &= 2u du \Rightarrow dx = \frac{2u}{u^2-1} du \end{align} \]
\begin{align*} \Rightarrow &\int \frac{1}{u} \cdot \frac{2u}{u^2-1} \, du \\[6pt] =& \int \frac{2}{(u-1)(u+1)} \, du \\[6pt] =& \int \left( \frac{1}{u-1} - \frac{1}{u+1} \right) du \\[6pt] =& \ln|u-1| - \ln|u+1| + C \\[6pt] =& \ln \left| \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1} \right| + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \frac{x^3 + 7x^2 + 18x + 17}{(x^2 + 4x + 5)^2} \, dx \)

\[ \begin{align*} &\frac{x^3 + 7x^2 + 18x + 17}{(x^2 + 4x + 5)^2} \\[6pt] =&\frac{Ax + B}{x^2 + 4x + 5} + \frac{Cx + D}{(x^2 + 4x + 5)^2} \\[6pt] \Rightarrow& A=1, B=3, C=1, D=2 \end{align*} \] \[ \Rightarrow \int \frac{x+3}{x^2+4x+5} \, dx + \int \frac{x+2}{(x^2+4x+5)^2} \, dx \\[6pt] \]
\[ \text{Let } u = x^2+4x+5, \quad du = (2x+4)dx \]
\begin{align*} &\Rightarrow \frac{1}{2} \int \frac{du}{u} + \int \frac{1}{(x+2)^2 + 1} \, dx + \frac{1}{2} \int \frac{du}{u^2} \\[6pt] &= \frac{1}{2} \ln|u| + \tan^{-1}(x+2) - \frac{1}{2u} + C \\[6pt] &=\frac{1}{2} \ln|x^2+4x+5| + \tan^{-1}(x+2) - \frac{1}{2x^2+8x+10} + C \end{align*}

\( \displaystyle \textbf{EX} \quad \int \frac{1}{\sqrt[3]{x} + \sqrt{x}} \, dx \)

\begin{align*} \text{Let } x &= u^6 \quad \Rightarrow \quad dx = 6u^5 \, du \\[6pt] \sqrt[3]{x} &= u^2, \quad \sqrt{x} = u^3 \end{align*}
\begin{align*} &\Rightarrow \int \frac{6u^5}{u^2 + u^3} \, du \\[6pt] &= \int \frac{6u^5}{u^2(1 + u)} \, du \\[6pt] &= \int \frac{6u^3}{u+1} \, du \\[6pt] &= 6 \int \left( u^2 - u + 1 - \frac{1}{u+1} \right) \, du \\[6pt] &= 6 \left( \frac{1}{3}u^3 - \frac{1}{2}u^2 + u - \ln|u+1| \right) + C \\[6pt] &= 2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} - 6\ln|\sqrt[6]{x} + 1| + C \end{align*}

8.8 Improper Intergrals

Improper Integrals with Infinite Limits

\( \displaystyle \textbf{Def} \quad \text{If a function} \, f \, \text{is continuous on the interval } \)

\[ \begin{align*} \int_a^{\infty} f(x) \, dx &= \lim_{b \to \infty} \int_a^b f(x) \, dx \\[6pt] \int_{-\infty}^b f(x) \, dx &= \lim_{a \to -\infty} \int_a^b f(x) \, dx \\[6pt] \int_{-\infty}^{\infty} f(x) \, dx &= \int_c^{\infty} f(x) \, dx + \int_{-\infty}^c f(x) \, dx \end{align*} \]

\( \displaystyle \textbf{EX} \quad \int_1^{\infty} e^{-x} \, dx \)

\[ \begin{align*} \int_1^{\infty} e^{-x} \, dx &= \lim_{b \to \infty} \int_1^b e^{-x} \, dx \\[6pt] &= \lim_{b \to \infty} \left[ -e^{-x} \right]_1^b \\[6pt] &= \lim_{b \to \infty} ( -e^{-b} - (-e^{-1}) ) \\[6pt] &= \lim_{b \to \infty} ( -e^{-b} + e^{-1} ) \end{align*} \]
\[ \begin{align*} b \to \infty, \quad e^{-b} = \frac{1}{e^b} \to 0 \end{align*} \]
\[ \begin{align*} &\Rightarrow 0 + e^{-1} \\[6pt] &= \frac{1}{e} \end{align*} \]

Improper Integrals with Infinite Discontinuities

不連續點在右端點 \( b \)

\[ \int_a^b f(x) \, dx = \lim_{c \to b^-} \int_a^c f(x) \, dx \]

不連續點在左端點 \( a \)

\[ \int_a^b f(x) \, dx = \lim_{c \to a^+} \int_c^b f(x) \, dx \]

不連續點在中間某點 \( c \)

\[ \int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx \]

ps. 兩個極限都收斂, 積分才算收斂

\( \displaystyle \textbf{EX} \quad \int_0^{\infty} \frac{1}{\sqrt{x}(x+1)} \, dx \)

\begin{align*} \text{Let } u &= \sqrt{x} \quad \Rightarrow \quad u^2 = x \\[6pt] dx &= 2u \, du \\[6pt] \text{Limits: } x \to 0 &\Rightarrow u \to 0 \\[6pt] x \to \infty &\Rightarrow u \to \infty \end{align*}
\[ \begin{align*} &\Rightarrow \int_0^{\infty} \frac{1}{u(u^2+1)} \cdot (2u \, du) \\[6pt] &= 2 \int_0^{\infty} \frac{1}{u^2+1} \, du \\[6pt] &= 2 \int_0^{\infty} \tan^{-1} u \, du \\[6pt] &= 2 \lim_{b \to \infty} \left[ \tan^{-1} u \right]_0^b \\[6pt] &= 2 \left( \lim_{b \to \infty} \tan^{-1} b - \tan^{-1} 0 \right) \\[6pt] &= 2 \left( \frac{\pi}{2} - 0 \right) \\[6pt] &= \pi \end{align*} \]

\( \displaystyle \textbf{EX} \quad \int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx \)

\[ \begin{align*} \int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx &= \int_{-\infty}^{0} \frac{1}{1+x^2} \, dx + \int_{0}^{\infty} \frac{1}{1+x^2} \, dx \\[6pt] &= \lim_{a \to -\infty} \int_{a}^{0} \frac{1}{1+x^2} \, dx + \lim_{b \to \infty} \int_{0}^{b} \frac{1}{1+x^2} \, dx \\[6pt] &= \lim_{a \to -\infty} \left[ \tan^{-1} x \right]_a^0 + \lim_{b \to \infty} \left[ \tan^{-1} x \right]_0^b \\[6pt] &= 0 - \left( -\frac{\pi}{2} \right) +\frac{\pi}{2} - 0 = \pi \end{align*} \]

\( \displaystyle \textbf{EX} \quad \int_{0}^{\pi} \sec x \, dx \)

\[ \begin{align*} \because& \, \sec x = \frac{1}{\cos x} \, \text{ ,when } x = \frac{\pi}{2} ,\, \cos \frac{\pi}{2} = 0 \\[6pt] \therefore& \, x = \frac{\pi}{2} \, \text{ isn't conti} \end{align*} \] \[ \begin{align*} \int_{0}^{\pi} \sec x \, dx &= \int_{0}^{\pi/2} \sec x \, dx + \int_{\pi/2}^{\pi} \sec x \, dx \\[6pt] &= \lim_{b \to \frac{\pi}{2}^-} \int_{0}^{b} \sec x \, dx + \lim_{c \to \frac{\pi}{2}^+} \int_{c}^{\pi} \sec x \, dx \end{align*} \] \[ \begin{align*} &\lim_{b \to \frac{\pi}{2}^-} \left[ \ln|\sec x + \tan x| \right]_{0}^{b} \\[6pt] &= \lim_{b \to \frac{\pi}{2}^-} \ln|\sec b + \tan b| - \ln|1 + 0| \\[6pt] &= \ln|\infty + \infty| - 0 \\[6pt] &= \infty \\[6pt] &\Rightarrow \int_{0}^{\pi} \sec x \, dx \, \text{ is div} \end{align*} \]

\( \displaystyle \textbf{EX} \quad \int_{-1}^{2} \frac{2}{x^3} \, dx \)

\begin{align*} \int_{-1}^{2} \frac{2}{x^3} \, dx &= \int_{-1}^{0} \frac{2}{x^3} \, dx + \int_{0}^{2} \frac{2}{x^3} \, dx \\[6pt] &= \lim_{b \to 0^-} \int_{-1}^{b} \frac{2}{x^3} \, dx + \lim_{c \to 0^+} \int_{c}^{2} \frac{2}{x^3} \, dx \end{align*}
\begin{align*} \lim_{c \to 0^+} \int_{c}^{2} 2x^{-3} \, dx &= \lim_{c \to 0^+} \left[ \frac{2x^{-2}}{-2} \right]_{c}^{2} \\[6pt] &= \lim_{c \to 0^+} \left[ -\frac{1}{x^2} \right]_{c}^{2} \\[6pt] &= \lim_{c \to 0^+} \left( -\frac{1}{4} + \frac{1}{c^2} \right) \\[6pt] &= \infty \end{align*}
\[ \Rightarrow \int_{-1}^{2} \frac{2}{x^3} \, dx \, \text{ diverges} \]

\( \displaystyle \textbf{EX} \quad \int_1^{\infty} xe^{-x} \, dx \)

\begin{align*} \text{Let } u &= x \quad \Rightarrow \quad du = dx \\[6pt] dv &= e^{-x} dx \quad \Rightarrow \quad v = -e^{-x} \\[6pt] \int xe^{-x} \, dx &= -xe^{-x} - \int (-e^{-x}) \, dx \\[6pt] &= -xe^{-x} - e^{-x} + C \end{align*}
\begin{align*} \int_0^{\infty} xe^{-x} \, dx &= \lim_{b \to \infty} \left[ -xe^{-x} - e^{-x} \right]_0^b \\[6pt] &= \lim_{b \to \infty} \left( \frac{-b}{e^b} - \frac{1}{e^b} \right) - \left( \frac{-1}{e^0} - \frac{1}{e^0} \right) \end{align*}
\begin{align*} &\because \lim_{b \to \infty} \frac{-b}{e^b} \quad (\text{Type } \infty/\infty) \\[6pt] &\xrightarrow{L'H} \lim_{b \to \infty} \frac{-1}{e^b} = 0 \\[6pt] &\therefore \int_0^{\infty} xe^{-x} \, dx = (0 - 0) - (-2) = 2 \end{align*}

Comparison Test

\( \displaystyle \text{If } f,g \text{ are continuous function and } f(x) \ge g(x) \ge 0 \quad \forall x \ge a \)
\[ \displaystyle \begin{align*} &\text{If } \int_a^{\infty} f(x) \, dx \text{ converges, then } \int_a^{\infty} g(x) \, dx \text{ converge} \\[6pt] &\text{If } \int_a^{\infty} g(x) \, dx \text{ diverges, then } \int_a^{\infty} f(x) \, dx \text{ diverge} \end{align*} \]

\( \displaystyle \textbf{EX} \quad \int_{e^e}^{\infty} \frac{1}{x \ln x [\ln(\ln x)]^2} \, dx \)

\begin{align*} u &= \ln x \implies du = \frac{1}{x} \, dx \\[6pt] \text{When } x &= e^e, \quad u = \ln(e^e) = e \\[6pt] \text{When } x &\to \infty, \quad u \to \infty \end{align*} \[ \Rightarrow \int_{e}^{\infty} \frac{1}{u (\ln u)^2} \, du \]
\begin{align*} \text{Let } v &= \ln u \implies dv = \frac{1}{u} \, du \\[6pt] \text{When }\ u &= e, \quad v = \ln(e) = 1 \\[6pt] \text{When } u &\to \infty, \quad v \to \infty \end{align*} \[ \Rightarrow \int_{1}^{\infty} \frac{1}{v^2} \, dv \]
\begin{align*} \int_{1}^{\infty} v^{-2} \, dv &= \lim_{b \to \infty} \int_{1}^{b} v^{-2} \, dv \\[6pt] &= \lim_{b \to \infty} \left[ -\frac{1}{v} \right]_{1}^{b} \\[6pt] &= \lim_{b \to \infty} \left( -\frac{1}{b} - (-1) \right) \\[6pt] &= 0 + 1 = 1 \end{align*}

Torricelli's Trumpet

\( \displaystyle y = \frac{1}{x} \, \text{在區間} \, [1, \infty) \, \text{繞x軸旋轉} \)

\[ \begin{align*} V &= \int_1^{\infty} \pi [R(x)]^2 \, dx \\[6pt] &= \lim_{b \to \infty} \int_1^b \pi \left( \frac{1}{x} \right)^2 \, dx \\[6pt] &= \pi \lim_{b \to \infty} \left[ -\frac{1}{x} \right]_1^b \\[6pt] &= \pi \lim_{b \to \infty} \left( -\frac{1}{b} + 1 \right) \\[6pt] &= \pi(0 + 1) = \pi \end{align*} \]
\[ \begin{align*} S &= \int_1^{\infty} 2\pi y \sqrt{1 + (y')^2} \, dx \\[6pt] &= \int_1^{\infty} 2\pi \left( \frac{1}{x} \right) \sqrt{1 + \left( -\frac{1}{x^2} \right)^2} \, dx \\[6pt] &= 2\pi \int_1^{\infty} \frac{\sqrt{1 + \frac{1}{x^4}}}{x} \, dx \end{align*} \] \[ \begin{align*} &\because x \ge 1 ,\quad \sqrt{1 + \frac{1}{x^4}} > 1 \\[6pt] &\therefore \frac{\sqrt{1 + \frac{1}{x^4}}}{x} > \frac{1}{x} \\[6pt] &\text{by C.T.} \quad \int_1^{\infty} \frac{1}{x} \, dx \quad \text{is div} \end{align*} \]

CH8 Caculus Test

\( \displaystyle \text{1.} \quad \int \tan x \ln(\cos x) \, dx \)
\[ \text{Let } u = \ln(\cos x) \quad du = \frac{-\sin x}{\cos x} dx = -\tan x \, dx \]
\[ \begin{align*} \int -u \, du &= -\frac{u^2}{2} + C \\[6pt] &= -\frac{(\ln(\cos x))^2}{2} + C \end{align*} \]
\( \displaystyle \text{2.} \quad \int \frac{1}{2+e^x} \, dx \)
\[ \begin{align*} \text{Let } \quad e^{x/2} &= \sqrt{2} \tan \theta \\[6pt] \frac{1}{2} e^{x/2} dx &= \sqrt{2} \sec^2 \theta \, d\theta \\[6pt] dx &= \frac{2\sqrt{2} \sec^2 \theta}{\sqrt{2} \tan \theta} \, d\theta \end{align*} \]
\[ \begin{align*} \Rightarrow &\int \frac{1}{2 \sec^2 \theta} \cdot \frac{2 \sqrt{2} \sec^2 \theta}{\sqrt{2} \tan \theta} d\theta \\[6pt] = &\int \cot \theta \, d\theta \\[6pt] = &\ln|\sin \theta| + C \\[6pt] = &\ln \left| \frac{e^{x/2}}{\sqrt{e^x+2}} \right| + C \end{align*} \]
\( \displaystyle \text{3.} \quad \int \ln(x + x^2) \, dx \) \[ \begin{align*} &\int \ln(x + x^2) \, dx \\[6pt] = &\int \ln x \, dx + \int \ln(1+x) \, dx \\[6pt] = &(x \ln x - x) + ((1+x) \ln(1+x) - (x+1)) + C \\[6pt] = & x(\ln x - 1) + (x+1)(\ln(x+1) - 1) + C \end{align*} \]
\( \displaystyle \text{4.} \quad \int e^{-3x} \sin(2x) \, dx \)
\[ \begin{align*} \text{Let } u &= \sin 2x \quad dv = e^{-3x}dx \\[6pt] du &=2\cos 2x \, dx \quad v=-\frac{1}{3}e^{-3x} \end{align*} \]
\[ \begin{align*} \int e^{-3x} \sin(2x) \, dx = -\frac{1}{3}e^{-3x} \sin(2x) + \frac{2}{3}\int e^{-3x}\cos 2x \end{align*} \]
\[ \begin{align*} \text{Let } u &= \cos 2x \quad dv = e^{-3x}dx \\[6pt] du &=-2\sin 2x \, dx \quad v=-\frac{1}{3}e^{-3x} \end{align*} \]
\[\begin{align*} &= -\frac{1}{3}e^{-3x} \sin(2x) + \frac{2}{3}\left( -\frac{1}{3}e^{-3x}\cos 2x - \frac{2}{3}\int e^{-3x} \sin(2x) \, dx \right) \\[6pt]&\Rightarrow \frac{13}{9}\int e^{-3x} \sin(2x) \, dx = -\frac{1}{3}e^{-3x} \sin(2x) -\frac{2}{9}e^{-3x} \cos(2x) \\[6pt]&\Rightarrow \int e^{-3x} \sin(2x) = -\frac{e^{-3x}}{13}\left( 3\sin 2x + 2\cos 2x \right) \end{align*}\]
\( \displaystyle \text{5.} \quad \int_{-\pi/2}^{\pi/2} \cos x \cos 7x \, dx \)
\[ \cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)] \]
\[ \begin{align*} &\Rightarrow \frac{1}{2} \int_{-\pi/2}^{\pi/2} (\cos 8x + \cos 6x) \, dx \\[6pt] &= \frac{1}{2} [\frac{1}{8}\sin 8x + \frac{1}{6}\sin 6x]|_{-\pi/2}^{\pi/2} \\[6pt] &=0 \end{align*} \]
\( \displaystyle \text{6.} \quad \int \cot^6(2x) \, dx \)
\[ u = 2x \quad dx = \frac{1}{2}du \quad cot^2 u = \csc^2 u - 1\]
\[\begin{align*} &\frac{1}{2} \int \cot^6 u \, du \\[6pt]=&\frac{1}{2} \int \cot^4 u (\csc^2 u - 1) \, du \\[6pt]=&\frac{1}{2} \left(\int \cot^4 u \csc^2 u \, du - \int \cot^4 u \, du \right) \\[6pt]=&\frac{1}{2} \left(\int \cot^4 u \csc^2 u \, du - \int \cot^2 u (\csc^2 u - 1) \, du \right) \\[6pt]=&\frac{1}{2} \left[\int \cot^4 u \csc^2 u \, du - \left(\int \cot^2 u \csc^2 u \, du - \int (\csc^2 u - 1) \, du\right) \right] \end{align*} \]
\[ \text{代換} \quad \frac{d}{du}(\cot u) = -\csc^2 u \text{ 與 } \int \cot^n u \csc^2 u \, du = -\frac{\cot^{n+1} u}{n+1} \]
\[\begin{align*} &\Rightarrow \frac{1}{2} \left[ -\frac{1}{5}\cot^5 u - \left( -\frac{1}{3}\cot^3 u - (-\cot u - u) \right) \right] + C \\[6pt] &= -\frac{1}{10}\cot^5(2x) + \frac{1}{6}\cot^3(2x) - \frac{1}{2}\cot(2x) - x + C \end{align*}\]
\(\displaystyle \text{7.} \quad \int \frac{\sqrt{9-x^2}}{x^2} \, dx ) \)
\[ x = 3\sin\theta \quad dx = 3\cos\theta d\theta \quad \cot\theta = \frac{\sqrt{9-x^2}}{x} \]
\[\begin{align*}&\int \frac{3\cos\theta}{9\sin^2\theta} \cdot 3\cos\theta \, d\theta \\[6pt]=& \int \cot^2\theta \, d\theta = \int (\csc^2\theta - 1) \, d\theta \\[6pt]=& -\cot\theta - \theta + C \\[6pt]=& -\frac{\sqrt{9-x^2}}{x} - \sin^{-1}\left(\frac{x}{3}\right) + C\end{align*}\]
\( \displaystyle \text{8.} \quad \int_{-1}^0 \frac{x^3}{x^2-2x+1} \, dx \)
\[ \frac{x^3}{(x-1)^2} = x+2 + \frac{3x-2}{(x-1)^2} = x+2 + \frac{3}{x-1} + \frac{1}{(x-1)^2} \]
\[\begin{align*} &\int_{-1}^0 \left( x+2 + \frac{3}{x-1} + \frac{1}{(x-1)^2} \right) \, dx\\[6pt] =&\left[ \frac{1}{2}x^2 + 2x + 3\ln|x-1| - \frac{1}{x-1} \right]_{-1}^0 \\[6pt]=&(0 + 0 + 3\ln 1 - (-1)) - \left( \frac{1}{2} - 2 + 3\ln 2 - (-\frac{1}{2}) \right) \\[6pt]=& 1 - ( -1 + 3\ln 2 ) \\[6pt]=& 2 - 3\ln 2 \end{align*}\]
\( \displaystyle \text{9.} \quad \int \frac{1}{\sqrt{x}-\sqrt[3]{x}+\sqrt[6]{x}-1} \, dx \)
\[ u = x^{1/6} \quad x = u^6 \quad dx = 6u^5 \, du \\[6pt] \frac{u^5}{u^3-u^2+u-1} = u^2+u+\frac{u^2}{u^3-u^2+u-1} \]
\[\begin{align*} &\int \frac{6u^5}{(u-1)(u^2+1)} \, du \\[6pt]=& \int \left( 6u^2 + 6u + \frac{3}{u-1} - \frac{3u}{u^2+1} + \frac{3}{u^2+1} \right) \, du \\[6pt]=& 2u^3 + 3u^2 + 3\ln|u-1| - \frac{3}{2}\ln(u^2+1) + 3\tan^{-1} u + C \\[6pt]=& 2x^{1/2} + 3x^{1/3} + 3\ln|x^{1/6}-1| - \frac{3}{2}\ln(x^{1/3}+1) + 3\tan^{-1}(x^{1/6}) + C \end{align*}\]
\( \displaystyle \text{10.} \quad \int_0^2 \frac{1}{\sqrt{|x-1|}} \, dx \)
\[ \int_0^2 f(x)dx = \lim_{t \to 1^-} \int_0^t \frac{1}{\sqrt{1-x}} dx + \lim_{s \to 1^+} \int_s^2 \frac{1}{\sqrt{x-1}} dx \]
\[\begin{align*} &\lim_{t \to 1^-} \left[ -2\sqrt{1-x} \right]0^t + \lim_{s \to 1^+} \left[ 2\sqrt{x-1} \right]_s^2 \\[6pt]=& \left( 0 - (-2\sqrt{1}) \right) + \left( 2\sqrt{1} - 0 \right) \\[6pt]=& 2 + 2 \\[6pt]=& 4 \end{align*}\]
\( \displaystyle \text{11.} \quad \int_0^{\infty} \frac{1}{(x+1)(x^2+1)} \, dx \)
\[ \text{部分分式: } \frac{1}{(x+1)(x^2+1)} = \frac{1/2}{x+1} - \frac{1/2(x-1)}{x^2+1} \quad \text{C.T.: } \frac{1}{(x+1)(x^2+1)} \ < \frac{1}{x^3} \]
\[\begin{align*} &\lim_{t \to \infty} \left[ \frac{1}{2}\ln|x+1| - \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\tan^{-1}x \right]_0^t \\[6pt]=& \lim_{t \to \infty} \left[ \frac{1}{4}\ln\left(\frac{(t+1)^2}{t^2+1}\right) + \frac{1}{2}\tan^{-1}t \right] - (0 + 0) \\[6pt]=& \frac{1}{4}\ln(1) + \frac{1}{2}\left(\frac{\pi}{2}\right) \\[6pt]=& \frac{\pi}{4} \end{align*}\]

9.1 Sequences

數列極限定義

\( \textbf{Def} \quad \text{Let } L \in \mathbb{R} \text{ The limit of a sequence } \{a_n\} \text{ is } L, \text{ written as:} \)

\[ \lim_{n \to \infty} a_n = L \]

\( \text{if for each } \epsilon > 0, \text{ there exists } M > 0 \text{ such that } |a_n - L| < \epsilon \text{ whenever } n > M. \)


** 極限存在則數列收斂至$L$, 不存在則發散 **

數列極限性質

\( \displaystyle \text{Let } \lim_{n \to \infty} a_n = L \text{ and } \lim_{n \to \infty} b_n = K. \)

\[ \begin{align*} \lim_{n \to \infty} (ca_n) &= cL, c \in \mathbb{R} \\[6pt] \lim_{n \to \infty} (a_n \pm b_n) &= L \pm K \\[6pt] \lim_{n \to \infty} (a_n b_n) &= LK \\[6pt] \lim_{n \to \infty} \frac{a_n}{b_n} &= \frac{L}{K}, b_n \neq 0, K \neq 0 \end{align*} \]

數列的夾擠定理(Squeeze Theorem)

\( \displaystyle \textbf{Thm} \quad \text{If } \lim_{n \to \infty} a_n = L = \lim_{n \to \infty} b_n \text{ and let } k \in \mathbb{N} , \quad \forall n \ge K \)

\( \text{such that } a_n \le c_n \le b_n \text{, then:} \)

\[ \lim_{n \to \infty} c_n = L \]

\( \displaystyle \textbf{EX} \quad \text{Show that } \lim_{n \to \infty} \frac{n!}{n^n} = 0 \)

\[ \begin{align*} 0 \le \frac{n!}{n^n} &= \frac{1 \cdot 2 \cdot 3 \cdots n}{n \cdot n \cdot n \cdots n} \\\\[6pt] &= \frac{1}{n} \left( \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n}{n} \right) \\\\[6pt] &\le \frac{1}{n} (1) = \frac{1}{n} \end{align*} \]
\[ \because \lim_{n \to \infty} 0 = 0 \quad \text{and} \quad \lim_{n \to \infty} \frac{1}{n} = 0 \] \[ \therefore \text{By Squeeze Theorem, } \lim_{n \to \infty} \frac{n!}{n^n} = 0 \]

Monotone Converge Theorem (MCT)

\[\text{If a sequence } \{a_n\} \text{ is a nodecreasing(increasing) sequence and is bounded above}\] \[ (a_n \le a_{n+1}, \, \forall n \in \mathbb{N})\]\[ \text{ then } \{a_n\} \text{ converges} \]

** 若數列遞增且有上界, 則為遞增 **

9.2 Series and Convergence

無窮級數定義

\( \textbf{Def} \quad \text{If } \{a_n\} \text{ is an infinite sequence, then the sum of its terms:} \)

\[ \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \dots + a_n + \dots \]

\( \text{is an infinite series}. \text{ The nth partial sum is given by:} \)

\[ S_n = a_1 + a_2 + \dots + a_n \]

\( \text{If the sequence of partial sums } \{S_n\} \text{ converges to } S, \text{ the series converges}. \)
\( \text{If } \{S_n\} \text{ diverges, the series diverges}. \)

Geometric Series (幾何級數)

\( \textbf{Def} \quad \text{A geometric series with ratio } r \text{ is of the form:} \)

\[ \sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \dots, \quad a \neq 0 \]

\( \textbf{Convergence of Geometric Series:} \)

\[ \begin{align*} &\text{1. If } |r| \lt 1, \text{ the series converges to } S = \frac{a}{1-r} \\\\[6pt] &\text{2. If } |r| \ge 1, \text{ the series diverges.} \end{align*} \]

\( \displaystyle \textbf{EX} \quad \text{Evaluate } \sum_{n=1}^{\infty} 3\left(\frac{1}{2}\right)^n \)

\[ \begin{align*} \text{First term } a &= 3\left(\frac{1}{2}\right)^1 = \frac{3}{2} \\[6pt] r &= \frac{1}{2} \\[6pt] \because |r| \lt 1, \quad S &= \frac{3/2}{1 - 1/2} = \frac{3/2}{1/2} = 3 \end{align*} \]

無窮級數的性質

\( \text{If } \sum a_n = A \text{ and } \sum b_n = B \text{ are convergent series, then:} \)

\[ \begin{align*} &\sum_{n=1}^{\infty} ca_n = cA \\\\[6pt] &\sum_{n=1}^{\infty} (a_n \pm b_n) = A \pm B \end{align*} \]

nth-Term Test(項測試法)

\( \displaystyle \textbf{Thm} \quad \text{If } \sum_{n=1}^{\infty} a_n \text{ converges, then } \lim_{n \to \infty} a_n = 0. \)

** 反過來不一定成立 **

nth-Term Test for Divergence

\[ \text{If } \lim_{n \to \infty} a_n \neq 0, \text{ then the series } \sum_{n=1}^{\infty} a_n \textbf{ diverges}. \]

** 若$P$則$Q$ $\iff$ 若非$Q$則非$P$ **

\( \displaystyle \textbf{EX} \quad \text{Does the series } \sum_{n=1}^{\infty} \frac{n}{2n+1} \text{ converge?} \)

\[ \begin{align*} \lim_{n \to \infty} a_n &= \lim_{n \to \infty} \frac{n}{2n+1} = \frac{1}{2} \\\\[6pt] \because \lim_{n \to \infty} a_n &\neq 0, \quad \therefore \text{By nth-term test, series diverges.} \end{align*} \]

9.3 The Intergral Test and p-Series

The Integral Test (積分測試法)

\( \begin{align*} \textbf{Thm} \quad \text{If }& f \text{ is positive, continuous, and decreasing} \\[6pt] &\text{ for } x \ge 1 \text{ and } a_n = f(n), \text{ then:} \end{align*} \)

\[ \sum_{n=1}^{\infty} a_n \text{ and } \int_1^{\infty} f(x) \, dx \]

\[ \text{Either both converge or both diverge.}\]

** \( \text{即 } f(x) \gt 0 \text{ and } f'(x) \lt 0 \) **

\( \displaystyle \textbf{EX} \quad \text{Determine the convergence of } \sum_{n=1}^{\infty} \frac{n}{n^2+1} \)

\( \displaystyle \text{Let } f(x) = \frac{x}{x^2+1}. \text{ Since } f \text{ is positive, continuous, and decreasing for } x \ge 1: \)

\[ \begin{align*}\int_1^{\infty} \frac{x}{x^2+1} , dx &= \lim_{b \to \infty} \int_1^b \frac{x}{x^2+1} , dx \\[6pt]&= \lim_{b \to \infty} \left[ \frac{1}{2} \ln(x^2+1) \right]1^b \\[6pt] &= \frac{1}{2} \lim_{b \to \infty} [ \ln(b^2+1) - \ln 2 ] = \infty\end{align*} \]

\[ \text{** 由積分發散推得級數發散 **}\]

p-Series and Harmonic Series(p級數與調和級數)

\( \textbf{Def} \quad \text{A series of the form:} \)

$$ \sum_{n=1}^{\infty} \frac{1}{n^p} = \frac{1}{1^p} + \frac{1}{2^p} + \frac{1}{3^p} + \dots $$

\[ \text{is called a p-series.} \]

\[ \begin{align*}&\text{ Conv. if } p \gt 1 \\[6pt]&\text{Div. if } 0 \lt p \le 1\end{align*} \]

\( \text{Special Case: } p=1 \text{ is the Harmonic Series} \sum \frac{1}{n}, \text{Div.} \)

\( \displaystyle \textbf{EX} \quad \text{Determine convergence} \)

\( \displaystyle \text{(a) } \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \implies p = \frac{1}{2} \le 1 \quad \therefore \textbf{Diverges} \)

\( \displaystyle \text{(b) } \sum_{n=1}^{\infty} \frac{1}{n^2} \implies p = 2 \gt 1 \quad \therefore \textbf{Converges} \)

9.4 Comparisons of Series

Direct Comparison Test (C.T.)

\( \textbf{Thm} \quad \text{Let } 0 < a_n \le b_n \text{ for all } n \)

\( \displaystyle \text{If } \sum_{n=1}^{\infty} b_n \text{ converges}, \text{ then } \sum_{n=1}^{\infty} a_n \text{ converges} \)

\( \displaystyle \text{If } \sum_{n=1}^{\infty} a_n \text{ diverges}, \text{ then } \sum_{n=1}^{\infty} b_n \text{ diverges} \)

\[ \because \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n \text{ is a convergent geometric series } (|r| < 1) \]\[ \therefore \sum_{n=1}^{\infty} \frac{1}{2 + 3^n} \text{ converges by C.T.} \]

\( \displaystyle \textbf{EX} \quad \text{Determine the convergence of } \sum_{n=1}^{\infty} \frac{1}{2 + 3^n} \)

\( \text{For } n \ge 1\)

\[ 2 + 3^n \gt 3^n \]\[ \Rightarrow \frac{1}{2 + 3^n} \lt \frac{1}{3^n} \]

\( \displaystyle \textbf{EX} \quad \text{Determine the convergence of } \sum_{n=3}^{\infty} \frac{\ln n}{n}\)

\( \text{For } n \ge 3 \)

\[\ln n \gt 1 \Rightarrow \frac{\ln n}{n} \gt \frac{1}{n}\] \[\begin{align*} \because \sum_{n=3}^{\infty} \frac{1}{n} \text{ is a divergent harmonic series} \\[6pt] \therefore \text{By C.T. } \quad \sum_{n=3}^{\infty} \frac{\ln n}{n} \textbf{diverges} \end{align*} \]

Limit Comparison Test (L.C.T.)

\( \textbf{Thm} \quad \text{Suppose} \quad a_n > 0 \quad b_n > 0, \text{ and } \)

\[ \lim_{n \to \infty} \frac{a_n}{b_n} = L \]

\( \text{If } 0 \lt L \lt \infty \text{ then:} \) \[\text{Both series} \sum{a_n} \text{ and } \sum{b_n} \text{ either both conv. or both div.}\]

\( \displaystyle \textbf{EX} \quad \text{Determine the convergence of } \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2 + 1} \)

\( \displaystyle \text{Let } a_n = \frac{\sqrt{n}}{n^2 + 1} \quad b_n = \frac{1}{n^{3/2}} \)

\[ \begin{align*}\lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \left( \frac{\sqrt{n}}{n^2 + 1} \cdot \frac{n^{3/2}}{1} \right) \\[6pt] &= \lim_{n \to \infty} \frac{n^2}{n^2 + 1} = 1\end{align*} \]
\[ \because \sum \frac{1}{n^{3/2}} \text{ is a convergent p-series } (p = \frac{3}{2} \gt 1) \]\[ \therefore \text{By L.C.T.} \quad \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2 + 1} \textbf{ conv.} \]

9.5 Alternating Series

The Alternating Series Test(A.S.T)

\( \text{Thm — The alternating series converges if the following two conditions are met:} \)

\( \text{1. } \lim_{n \to \infty} a_n = 0 \)

\( \text{2. } a_{n+1} \le a_n \text{ for all } n \text{ (Monotonic decreasing).} \)

3. Absolute and Conditional Convergence

\( \text{For a series } \sum a_n: \)

  • Absolute Convergence: \( \text{If } \sum |a_n| \text{ converges.} \)
  • Conditional Convergence: \( \text{If } \sum a_n \text{ converges, but } \sum |a_n| \text{ diverges.} \)

\( \text{Thm — If } \sum |a_n| \text{ converges, then } \sum a_n \text{ converges.} \)

\( \displaystyle \text{EX — Classify } \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \)

\[ \begin{align*} &\text{1. By AST, the series converges.} \\ &\text{2. Check absolute values: } \sum \left| \frac{(-1)^{n-1}}{n} \right| = \sum \frac{1}{n} \quad (\text{Divergent harmonic series}) \end{align*} \]
\[ \therefore \text{The series is conditionally convergent.} \]

4. Alternating Series Remainder

\( \text{If an alternating series converges to } S, \text{ the remainder } |R_n| = |S - S_n| \text{ satisfies:} \)

\[ |R_n| \le a_{n+1} \]

\( \text{The error is no greater than the first neglected term.} \)

9.6 The Ratio and Root Tests

The Ratio Test (比值審斂法)

\( \textbf{Thm} \quad \text{Let } \sum a_n \text{ be a series with nonzero terms.} \) \[ \text{Let } \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = r \] \[ \begin{align*} &\text{If } r \lt 1, \text{ the series converges absolutely.} \\[6pt] &\text{If } r \gt 1 \text{ or } r = \infty, \text{ the series diverges.} \\[6pt] &\text{If } r = 1, \text{ the test is inconclusive.} \end{align*} \]

** 常用於含階乘 (n!) 或指數項 (an) 的級數 **

The Root Test (根值審斂法)

\( \textbf{Thm} \quad \text{Let } \sum a_n \text{ be a series.} \) \[ \text{Let } \lim_{n \to \infty} \sqrt[n]{|a_n|} = r \] \[ \begin{align*} &\text{If } r \lt 1, \text{ the series converges absolutely.} \\[6pt] &\text{If } r \gt 1 \text{ or } r = \infty, \text{ the series diverges.} \\[6pt] &\text{If } r = 1, \text{ the test is inconclusive.} \end{align*} \]

9.7 Taylor Polynomials and Approximations

Taylor Polynomials (泰勒多項式)

\( \textbf{Def} \quad \text{If } f \text{ has } n \text{ derivatives at } c, \text{ then the nth Taylor polynomial for } f \text{ at } c \text{ is:} \) \[ P_n(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \dots + \frac{f^{(n)}(c)}{n!}(x-c)^n \]

** 若中心點 ( c = 0 ), 則稱為麥克勞林多項式 (Maclaurin Polynomial) **

Taylor's Theorem & Remainder (泰勒定理與餘項)

\( \textbf{Thm} \quad \text{若 } f(x) = P_n(x) + R_n(x), \text{ 則餘項的拉格朗日形式為：} \) \[ R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1} \] \( \text{其中 } z \text{ 介於 } x \text{ 與 } c \text{ 之間。} \)

9.8 Power Series

冪級數定義

\( \textbf{Def} \quad \text{以 } c \text{ 為中心的冪級數形式為：} \) \[ \sum_{n=0}^{\infty} a_n(x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \dots \]

收斂半徑與區間 (Radius and Interval of Convergence)

\( \textbf{Thm} \quad \text{對於冪級數，其收斂情形必為下列之一：} \) \[ 僅在中心點收斂：收斂半徑 ( R = 0 )。\] \[在有限區間內收斂：存在 ( R \gt 0 )，使得當 ( |x-c| \lt R ) 時絕對收斂。\] \[在全體實數收斂：收斂半徑 ( R = \infty )。\]

冪級數的微分與積分

\( \textbf{Thm} \quad \text{在收斂區間內，冪級數可逐項微分與積分：} \) \[ \begin{align*} f'(x) &= \sum_{n=1}^{\infty} n \, a_n (x-c)^{n-1} \\[6pt] \int f(x) \, dx &= C + \sum_{n=0}^{\infty} a_n \frac{(x-c)^{n+1}}{n+1} \end{align*} \]

9.9 Representation of Functions by Power Series

幾何冪級數 (Geometric Power Series)

\[ \frac{a}{1-r} = \sum_{n=0}^{\infty} ar^n, \quad |r| \lt 1 \]
\( \displaystyle \textbf{EX} \quad \text{Find a series for } f(x) = \frac{1}{1-x} \text{ centered at 0:} \) \[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots = \sum_{n=0}^{\infty} x^n, \quad |x| \lt 1 \]

9.10 Taylor and Maclaurin Series

Taylor Series

\( \textbf{Def} \quad \text{若函數 } f \text{ 在 } c \text{ 處具有各階導數，其級數展開為：} \) \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n \]

Maclaurin series of common basic functions

函數 麥克勞林級數展開式 收斂區間
\[ e^x \] \[ \displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \dots \] \[ (-\infty, \infty) \]
\[ \sin x \] \[ \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \dots \] \[ (-\infty, \infty) \]
\[ \cos x \] \[ \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \dots \] \[ (-\infty, \infty) \]
\[ \frac{1}{1-x} \] \[ \displaystyle \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \dots \] \[ (-1, 1) \]
\[ \ln(x+1) \] \[ \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \] \[ (-1, 1] \]
\[ \arctan x \] \[ \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots \] \[ [-1, 1] \]