微積分 - 積分¶

積分Intergration¶

積分是用來求曲線下面積的一種工具

定積分¶

定義：
設\(y=f(x)\)在閉區間\([a,b]\)上連續
\(S\)是\([a,b]\)的任意一種分割
得到\(n\)個區間\([x_{k-1},x_k],k\in\mathbb{N}\cap(0,n]\)
且\(a=x_0<x_1<\dots<x_{n-1}<x_n=b\)
若於每一區間上取一\(\varepsilon_k\)
使得\(x_k\le\varepsilon_k\le{x_{k+1}}\)
再取一\(\delta=\max\{x_1-x_0,x_2-x_1,x_3-x_2\dots,x_{n}-x_{n-1}\}\)
則定義
\(\displaystyle\lim_{\delta\to{0}}\sum_{k=0}^{n}f(\varepsilon_{k})(x_{k+1}-x_k)=\displaystyle\lim_{\delta\to{0}}\sum_{k=0}^{n}f(\varepsilon_{k})(\Delta{x_{k}})=\displaystyle\int_{a}^{b}f(x)dx\)

為何\(\delta=\max\{x_1-x_0,x_2-x_1,x_3-x_2\dots,x_{n}-x_{n-1}\}?\)
答：若整個分割中最大塊的區間長度趨近於0，那麼其餘的分割區間長度也必趨近於0。

等分¶

由定積分的嚴格定義，可取將\([a,b]\ n\)等分的分割\(S\)
即\(|[x_{k-1},x_k]|=|[x_k,x_{k+1}]|,\forall k\in\mathbb{N}\)

可得\(\displaystyle\sum_{k=1}^nf(\dfrac{k}{n})\dfrac{b-a}{n}\approx\int_{a}^{b}f(x)dx\)

而由定義，當\(\delta\to 0\)時，\(n\to\infty\)
故\(\displaystyle\lim_{n\to\infty}\sum_{k=1}^nf(\dfrac{k}{n})\dfrac{b-a}{n}=\int_{a}^{b}f(x)dx\)

上和與下和：
若\(f(x)\)在\([a,b]\)遞增

\(\displaystyle\sum_{k=1}^nf(\dfrac{k}{n})\dfrac{b-a}{n}\)稱為上和\(M\)

\(\displaystyle\sum_{k=0}^{n-1}f(\dfrac{k}{n})\dfrac{b-a}{n}\)稱為下和\(m\)

微積分基本定理¶

反導函數：若\(F'(x)=f(x)\)，則稱\(F(x)\)為\(f(x)\)的一個反導函數

若\(F'(x)=f(x)\)，則\(F(x)+c=\displaystyle\int f(x)dx\)，\(c\)是一個常數
\(若F(x)\)是\(f(x)\)的一個反導函數，則\(\displaystyle\int_a^bf(x)dx=F(b)-F(a)\)

\(pf:\)
先證明2.
\(\because F(x)\)是\(f(x)\)的一個反導函數\(\therefore F'(x)=f(x)\)
而\(\displaystyle\int_{a}^bf(x)dx=\lim_{\delta \to 0}\sum_{k=1}^{n}f(\varepsilon_k)(x_{k+1}-x_k)\)

又由均值定理：若\(f(x)\)在\((a,b)\)上連續，則\(\exists c\in(a,b)\ s.t.\ f'(c)=\dfrac{f(b)-f(a)}{b-a}\)

而此時對於區間\([x_{k},x_{k+1}]\)，也有一常數\(c\in(x_{k},x_{k+1})\)使得\(F'(c)=\dfrac{F(x_{k+1})-F(x_k)}{x_{k+1}-x_k}\)，則此時取\(\varepsilon_k\)為此常數，則

\(\displaystyle\int_{a}^{b}f(x)dx=\lim_{\delta\to 0}\sum^{n}_{k=1}f(\varepsilon_k)(x_{k+1}-x_k) \\=\displaystyle\lim_{\delta\to 0}\sum^{n}_{k=1}F'(\varepsilon_k)(x_{k+1}-x_k) \\=\displaystyle\lim_{\delta\to 0}\sum^{n}_{k=1}\dfrac{F(x_{k+1})-F(x_k)}{x_{k+1}-x_k}(x_{k+1}-x_k)=\lim_{\delta\to 0}\sum^{n}_{k=1}[F(x_{k+1})-F(x_k)]=F(b)-F(a)\)

再證明1.
因為已知\(\displaystyle\int_a^bf(x)dx=F(b)-F(a)\)
因此\(\displaystyle\int^{x}_{a}f(x)dx=F(x)-F(a)\)
而當下界也不確定時，即\(\displaystyle\int f(x)dx=F(x)+C\)
微積分基本定理從而得證。

運算性質¶

\(\displaystyle\int_{a}^{b}f(x)dx=\int^c_af(x)dx+\int^b_cf(x)dx\)
\(\displaystyle\int_{a}^{b}f(x)dx=-\int_b^af(x)dx\)
若\(f(x)\)為奇函數，則\(\displaystyle\int^a_{-a}f(x)dx=0\)。
若\(f(x)\)為偶函數，則\(\displaystyle\int^a_{-a}f(x)dx=2\int^a_{0}f(x)dx\)
\(\displaystyle\int_a^af(x)dx=0\)
\(\displaystyle\int cf(x)dx=c\int f(x)dx\)
\(\displaystyle\int [f(x)\pm g(x)]dx=\int f(x)dx\pm\int g(x)dx\)
\(\displaystyle\int f(x)dx=\int f(u)du=\int f(t)dt\)
\(\displaystyle\int_a^b f(x)dx=\int_b^a f(a+b-x)dx\)

積分方法¶

變數代換¶

設\(F(x)\)為\(f(x)\)的一個反導函數，且\(g(x)\)可微

\(\because(F(g(x)))'=F'(g(x))g'(x)=f(g(x))g'(x) \\\therefore\displaystyle\int f(g(x))g'(x)dx=F(g(x))\)

而因為可令\(u=g(x)\)，則\(du=g'(x)dx\)，故稱為變數代換

例如
\(\displaystyle\int\dfrac{x}{\sqrt{1-x^2}}dx \\\overset{u=1-x^2}{=}\displaystyle\dfrac{1}{2}\int\dfrac{du}{\sqrt{u}}=\dfrac{1}{2}(-2\sqrt{1-x^2})+\text{C}=-\sqrt{1-x^2}+\text{C}\)

分部積分¶

由微分乘則\((f(x)g(x))'=f'(x)g(x)+f(x)g'(x)\)
可得\(\displaystyle\int(f(x)g(x))'dx=\int [f'(x)g(x)+f(x)g'(x)]dx\)
\(\implies f(x)g(x)=\displaystyle\int f(x)dg(x)+\int g(x)df(x) \\\therefore\displaystyle\int f(x)dg(x)=f(x)g(x)-\int g(x)df(x)\)

例如
\(\displaystyle\int e^x\cos{x}dx\)
令\(dv=e^xdx,u=\cos{x}\)
\(\therefore du=-\sin{x}dx,v=e^x\)
\(\implies\displaystyle\int e^x\cos{x}dx=e^x\cos{x}+\int e^x\sin{x}dx \\=\displaystyle e^x\cos{x}+(e^x\sin{x}-\int e^{x}\cos{x}dx) \\\implies\displaystyle\int e^x\cos{x}dx=\dfrac{e^{x}(\cos{x}+\sin{x})}{2}+\text{C}\)

\(DI\ Method\)

三角代換¶

使用時機

看到\(\sqrt{a^2-x^2}\)令\(x=a\sin{u}\)
看到\(\sqrt{x^2-a^2}\)令\(x=a\sec{u}\)
看到\(a^2+x^2\)令\(x=a\tan{u}\)

例如
\(\displaystyle\int\sqrt{1-x^2}dx\)
令\(x=\sin{u},dx=\cos{u}du\)
則原式
\(\displaystyle\int\cos^2{u}du=\int\dfrac{1+\cos{2u}}{2}du=\dfrac{u}{2}+\dfrac{\sin{2u}}{4}+\text{C}=\dfrac{\sin^{-1}{x}}{2}-\dfrac{x\sqrt{1-x^2}}{2}+\text{C}\)

部分分式¶

當遇到有理函式時，如\(\dfrac{p(x)}{q(x)}\)，其中\(p(x),q(x)\)皆為實係數多項式，則可使用此積分方法

\(\dfrac{A}{x-a}\)
\(\dfrac{A}{(x-a)^n}\)
\(\dfrac{Cx+D}{(x^2+px+q)^n}\)

\(n\in\mathbb{N},p^2-4q<0\)

例如：
\(\displaystyle\int\dfrac{1}{x^2-1}dx=?\)
令\(\dfrac{1}{(x+1)(x-1)}=\dfrac{A}{x+1}+\dfrac{B}{x-1}\)
則\(1=A(x-1)+B(x+1)\)
故\(\begin{cases}A=\dfrac{-1}{2}\\B=\dfrac{1}{2}\end{cases}\)
因此\(\displaystyle\int\dfrac{1}{x^2-1}dx=\int\dfrac{-1}{2(x+1)}dx+\int\dfrac{1}{2(x-1)}dx=\dfrac{1}{2}(\ln|\dfrac{x-1}{x+1}|)+\text{C}=-\tanh^{-1}x+\text{C}\)

基本函數積分公式¶

冪函數¶

\(\displaystyle\int x^{p}dx=\dfrac{x^{p+1}}{p+1}+\text{C},p\in\mathbb{R}-\{-1\}\)

\(\displaystyle\int\dfrac{1}{x}dx=\ln|x|+\text{C}\)

指對數¶

\(\displaystyle\int e^{cx}dx=\dfrac{e^{cx}}{c}+\text{C},c\in\mathbb{R}-\{0\}\)

\(\displaystyle\int\log_axdx=\dfrac{x\ln{x}-x}{\ln{a}}+\text{C}\)

三角函數¶

由微積分基本定理能夠輕易得出正餘弦函數的不定積分

\(\because\dfrac{d}{dx}\sin{x}=\cos{x},\dfrac{d}{dx}\cos{x}=-\sin{x}\)
\(\implies\displaystyle\int\cos{x}dx=\sin{x}+c\)
以及\(\displaystyle\int\sin{x}dx=-\cos{x}+c\)

正餘弦函數積分相當於圖形向右平移\(\dfrac{\pi}{2}\)後，在\(y\)方向加上某個常數。
但正餘切和正餘割函數則較複雜

\(\displaystyle\int\tan{x}dx=-\ln{|\cos{x}|}+c\)
\(\displaystyle\int\sec{x}dx=\ln{|\tan{x}+\sec{x}|}+c\)

\(pf: \\\displaystyle\int\tan{x}dx \\=\displaystyle\int\dfrac{\sin{x}}{\cos{x}}dx\)
令
\(u=\cos{x}\)
\(\therefore du=-\sin{x}dx\)

\(\displaystyle\int\tan{x}dx \\=\displaystyle\int\dfrac{\sin{x}}{\cos{x}}\dfrac{du}{-\sin{x}} \\=\displaystyle\int\dfrac{-1}{u}du \\=-\ln{|u|} \\=-\ln{|\cos{x}|}+c(=\ln{|\sec{x}|})\)

法一：

\(\displaystyle\int\sec{x}dx \\=\displaystyle\int\dfrac{\sec{x}(\sec{x}+\tan{x})}{(\sec{x}+\tan{x})}dx \\=\displaystyle\int\dfrac{\sec^2{x}+\sec{x}\tan{x}}{\sec{x}+\tan{x}}dx\)

令\(u=\sec{x}+\tan{x}\therefore du=\tan{x}\sec{x}+\sec^2{x}\)

\(\therefore\displaystyle\int\dfrac{\sec^2{x}+\sec{x}\tan{x}}{\sec{x}+\tan{x}}dx=\displaystyle\int\dfrac{du}{u}=\ln{|u|}=\ln{|\sec{x}+\tan{x}|}+c\)

法二：
\(\displaystyle\int\sec{x}dx \\=2\displaystyle\int\dfrac{1+t^2}{1-t^2}\dfrac{dt}{1+t^2}\)
\(t=\tan{\dfrac{x}{2}}\therefore dx=\dfrac{2dt}{1+t^2}\)

\(\therefore\displaystyle\int\sec{x}dx=\ln{|\dfrac{1+t}{1-t}|}+C \\=\ln{|\dfrac{(t+1)^2}{1-t^2}|}+C=\ln{|\dfrac{1+\sin{x}}{\cos{x}}|}+C=\ln{|\sec{x}+\tan{x}|}+C\)

高次三角函數與其高次乘積¶

\(\displaystyle\int\sin^{n}xdx=\dfrac{1}{n}(-\sin^{n-1}x\cos{x}+(n-1)\int\sin^{n-2}xdx)\)
\(\displaystyle\int\sec^{n}xdx=\dfrac{1}{n-1}(\sec^{n-2}x\tan{x}+(n-2)\int\sec^{n-2}xdx)\)
\(\csc^{n}x\)

只證明1.其餘可利用類似想法完成
皆為先分部積分再利用平方關係化為遞迴關係。
\(\displaystyle\int\sin^{n}xdx=-\int\sin^{n-1}xd\cos{x}=-\sin^{n-1}x\cos{x}+\int\cos{x}d\sin^{n-1}x\)
又
\(d\sin^{n-1}x=(n-1)\sin^{n-2}x\cos{x}dx\)
\(\displaystyle\therefore\int\sin^{n}xdx\\=\displaystyle-\sin^{n-1}x\cos{x}+(n-1)\int\cos^2{x}\sin^{n-2}xdx\\=\displaystyle-\sin^{n-1}x\cos{x}+(n-1)\int(1-\sin^2{x})\sin^{n-2}xdx\\=\displaystyle-\sin^{n-1}x\cos{x}+(n-1)(\int\sin^{n-2}xdx-\int\sin^{n}xdx) \\\implies\displaystyle\int\sin^{n}xdx=\dfrac{1}{n}(-\sin^{n-1}x\cos{x}+(n-1)\int\sin^{n-2}xdx)\)

其餘可利用平方關係與以下公式化簡

\(\sin^{n}x\cos^{m}x\)

先討論奇偶
一奇一偶奇減一
二偶用平方換
二奇 cos減一
次方相同時二倍角
\(\displaystyle\int\sin^{2n+1}x\cos^{2m}xdx=-\int\sin^{2n}x\cos^{2m}xd\cos{x}=-\int(1-\cos^{2}x)^n\cos^{2m}xd\cos{x}\)
令\(u=\cos{x}\)，即求\(\displaystyle\int(1-u^2)^nu^{2m}du\)
以二項式定理展開後得：
\(-\displaystyle\int[\sum_{k=0}^{n}(-1)^k\binom{n}{k}u^{2m+2k}]du \\=\displaystyle\sum_{k=0}^{n}[(-1)^{k+1}\binom{n}{k}\int{u^{2m+2k}}du] \\=\displaystyle\sum_{k=0}^{n}[(-1)^{k+1}\binom{n}{k}\dfrac{\cos^{2m+2k+1}x}{2m+2k+1}]+C\)
\(\therefore\displaystyle\int\sin^{2n+1}x\cos^{2m}xdx=\sum_{k=0}^{n}[(-1)^{k+1}\binom{n}{k}\dfrac{\cos^{2m+2k+1}x}{2m+2k+1}]+C\)
\(\displaystyle\int\sin^{2n}x\cos^{2m}x=\int(1-\cos^{2}x)^n\cos^{2m}xdx=\int\sum_{k=0}^{n}(-1)^k\cos^{2k+2m}xdx\)
再使用高次三角函數漸化式化簡。
2. \(\sec^{n}x\tan^{m}x\)

\(\csc^{n}x\cot^{m}x\)

多種三角函數合在一起時可使用\(\tan{2x}\)萬能公式
化為部分分式再積分

其他雜項¶

\(\displaystyle\int x^ne^{mx}dx=\sum^{n}_{k=0}\dfrac{(-1)^{k}}{m^{k+1}}\binom{n}{k}k!x^{n-k}e^{mx}+C,n\in\mathbb{N},m\in\mathbb{R}\backslash\{0\}\)
\(\displaystyle\int\sin{ax}\cos{bx}dx=\dfrac{-1}{2}(\dfrac{\cos{((a+b)x)}}{a+b}+\dfrac{\cos{((a-b)x)}}{a-b})+C,a,b\in\mathbb{R}\land a^2+b^2\not=0\land a\not=b\)

積分應用¶

求弧長¶

一小段弧長可近似為直線，故\(dl=\sqrt{(dx)^2+(dy)^2}\)
故函數\(f(x)\)在區間\([a,b]\)之弧長為
\(\displaystyle\int_{a}^{b}dl=\int_{a}^{b}\sqrt{1+(\dfrac{dy}{dx})^2}dx\)

推廣：
參數式\(\begin{cases}x=x(t)\\y=y(t)\end{cases}\quad t\in[a,b]\)
其弧長為
\(\displaystyle\int_{a}^{b}\sqrt{(dx)^2+(dy)^2}=\int_{a}^{b}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}dt\)

求曲線下面積¶

函數\(f(x)\)在\([a,b]\)上連續，則其與\(x\)軸所圍成的面積為
\(\displaystyle\int^{b}_{a}|f(x)|dx\)

例如：
求\(f(x)=\sqrt{(x+1)(x-1)}\)與\(\begin{cases}(x-3)(x-6)=0\\y=0\end{cases}\)所圍成的圖形面積

先作圖，確定圖形形狀，上下界之間有根需分段積分，負區變號
what

再寫積分式

\(\displaystyle\int^{6}_{3}|\sqrt{x^2-1}|dx=\int^{6}_{3}\sqrt{x^2-1}dx\)

利用三角代換
令\(x=\sec{u},dx=\tan{u}\sec{u}du\)
則原式變為
\(\displaystyle\int^{6}_{3}\tan^2{u}\sec{u}du\)
先求其不定積分
\(\displaystyle\int\sec^{3}{u}-\sec{u}du=\int\sec^{3}{u}du-\int\sec{u}du\\ =\displaystyle\dfrac{1}{2}\tan{u}\sec{u}+\dfrac{1}{2}\int\sec{u}du-\int\sec{u}du \\=\dfrac{1}{2}\tan{u}\sec{u}-\dfrac{1}{2}\ln{|\sec{u}+\tan{u}|}+C\)

而因為\(x=\sec{u}\therefore\tan{u}=\sqrt{x^2-1}\)
因此\(\displaystyle\int\sqrt{x^2-1}dx=\dfrac{1}{2}(x\sqrt{x^2-1}-\ln{|x+\sqrt{x^2-1}|})+C\)

將上下界代入後
\(\displaystyle\int^{6}_{3}\sqrt{x^2-1}dx=3\sqrt{35}-3\sqrt{2}+\ln{\sqrt{\dfrac{3+2\sqrt{2}}{6+\sqrt{35}}}}\)即為所求

定積分求體積¶

若有一面積函數\(A(x)\)，則其與\((x-a)(x-b)=0\)兩平面所圍成的體積為
\(|\displaystyle\int^{b}_{a}A(x)dx|\)

例如：
求證三角錐體積\(V\)為\(\dfrac{A_{0}h}{3}\)，其中\(A_0\)，\(h\)分別為三角錐之底面積，高

\(pf:\)

what

求旋轉體體積¶

旋轉體體積定義：\(f(x)\)繞\(x\)軸旋轉所圍成的體積

圓盤法

利用定積分求體積的方法，先寫出面積函數，再做積分
面積函數為每個圓盤的面積，即以\(|y|\)值為半徑的圓面積\(\pi{y^2}\)
則函數\(y=f(x)\)繞\(x\)軸旋轉與\((x-a)(x-b)=0\)兩平面所圍成之體積為
\(|\displaystyle\int_{a}^{b}\pi{y^2}dx|\)

套筒法

此法通常使用於\(y=f(x)\)繞\(y\)軸旋轉的體積
即求\(y=f(x)\)繞\(y\)軸旋轉與\((y-a)(y-b)=0\)所形成的體積
作法為：
先將每個小中空圓柱面的面積函數算出，再將這些小圓柱體面積加起來即為所求
故所求為\(|\displaystyle\int_{a}^{b}2\pi xydx|\)

例題：
試證明球體體積\(V_S=\dfrac{4}{3}\pi{R^3}\)，其中\(R\)為球半徑。

考慮平面上一半徑為\(R\)，以原點為圓心之圓，則其方程式為\(x^2+y^2=R^2\)
將上半圓繞\(x\)軸旋轉一圈所形成的旋轉體體積即為半徑為\(R\)之球體體積
因此所求\(V_S=\displaystyle\int_{-R}^{R}\pi(\sqrt{R^2-x^2})^2dx=2\int_{0}^{R}\pi(R^2-x^2)dx=\dfrac{4}{3}\pi{R^3}\)

變化型：

有兩函數\(f(x)\)與\(g(x)\)，求其繞\(x\)軸與\((x-a)(x-b)=0\)所形成的體積

解：
\(|\displaystyle\int_{a}^{b}\pi(f(x)-g(x))^2dx|\)

求旋轉體表面積¶

求\(f(x)\)繞\(x\)軸與\((x-a)(x-b)=0\)所形成的旋轉體表面積\(A_r\)
面積就是弧長的積分
因此先將弧長算出後再求出旋轉體積分
\(|\displaystyle\int_{a}^{b} 2\pi f(x) dl|\)
其中\(dl=\sqrt{1+(\dfrac{dy}{dx})^2}dx\)
故\(A_r=|\displaystyle\int_{a}^{b} 2\pi f(x) \sqrt{1+(f'(x))^2}dx|\)

例題：
試證明球體表面積\(A_S=4\pi R^2\)，其中\(R\)為球半徑。

考慮平面上一半徑為\(R\)，以原點為圓心之圓，則其方程式為\(x^2+y^2=R^2\)
將上半圓繞\(x\)軸旋轉一圈所形成的旋轉體表面積即為半徑為\(R\)之球體表面積
故所求\(A_S=\displaystyle\int_{-R}^{R}2\pi(\sqrt{R^2-x^2})(\sqrt{1+(\dfrac{-2x}{2\sqrt{R^2-x^2}})^2}dx\)
\(=\displaystyle 4\pi\int_{0}^{R}\sqrt{R^2-x^2}\sqrt{\dfrac{R^2-x^2+x^2}{R^2-x^2}}dx \\=\displaystyle 4\pi R\int_{0}^{R}dx=4\pi{R^2}\)

定積分例題¶

\(\displaystyle\int_{0}^{\pi}\dfrac{x\sin{x}}{1+\cos^{2}x}dx=?\)

\(\displaystyle\int_a^bf(x)dx=\int_b^a f(a+b-x)dx\)

令\(I=\displaystyle\int_{0}^{\pi}\dfrac{x\sin{x}}{1+\cos^{2}x}dx\)
\(\therefore I=\displaystyle\int_{\pi}^{0}\dfrac{(\pi-x)\sin{(\pi-x)}}{1+\cos^{2}(\pi-x)}dx=\pi\displaystyle\int_{\pi}^{0}\dfrac{\sin{(x)}}{1+\cos^{2}(x)}dx-I\)
\(2I=\pi(\tan^{-1}{(\cos{x})}\Bigg{|}_0^{\pi})=\dfrac{\pi^2}{2}\)
\(\implies I=\displaystyle\int_{0}^{\pi}\dfrac{x\sin{x}}{1+\cos^{2}x}dx=\dfrac{\pi^2}{4}\)

證明：
\(\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\sin{\theta})}d\theta=\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\cos{\theta})}d\theta=\dfrac{-\pi}{2}\ln2\)

\(\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\sin{\theta})}d\theta=\displaystyle\int^{0}_{\frac{\pi}{2}}\ln{(\cos{(\dfrac{\pi}{2}-\theta)})}d\theta=\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\cos{(\dfrac{\pi}{2}-\theta)})}d(\dfrac{\pi}{2}-\theta)=\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\cos{\theta})}d\theta\)

又\(\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\cos{\theta})}d\theta=\dfrac{1}{2}(\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\sin{\theta})}d\theta+\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\cos{\theta})}d\theta)\)
\(=\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\dfrac{1}{2}\sin{2\theta})}d2\theta=\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{\dfrac{1}{2}}d2\theta+\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\sin{2\theta})}d2\theta\)
\(=\dfrac{\pi}{4}\ln({\dfrac{1}{2}})+\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\sin{2\theta})}d2\theta\)

令\(I=\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\sin{\theta})}d\theta\)
則：\(I=\dfrac{\pi}{4}\ln({\dfrac{1}{2}})+\dfrac{1}{4}\displaystyle\int_{0}^{\frac{\pi}{2}}\ln{(\sin{2\theta})}d2\theta \\=\dfrac{\pi}{4}\ln({\dfrac{1}{2}})+\dfrac{1}{2}I\)

故\(I=\dfrac{-\pi}{2}\ln{2}\)

推廣：
設\(f(x)\)在\([0,\dfrac{\pi}{2}]\)區間上連續，則
\(\displaystyle\int_{0}^{\frac{\pi}{2}}f(\sin{x})dx=\displaystyle\int_{0}^{\frac{\pi}{2}}f(\cos{x})dx\)
利用變數代換以及三角恆等式
令\(u=\dfrac{\pi}{2}-x\)，則\(du=-dx\)
可得\(\displaystyle\int_{0}^{\frac{\pi}{2}}f(\sin{x})dx=-\displaystyle\int_{\frac{\pi}{2}}^{0}f(\sin{(\dfrac{\pi}{2}-u)})du=\displaystyle\int_{0}^{\frac{\pi}{2}}f(\cos{u})du=\displaystyle\int_{0}^{\frac{\pi}{2}}f(\cos{x})dx\)