第四章 不定積分（二）

  本文接自上一篇《第四章 不定積分（一）》，繼續記錄總習題四。

總習題四

4.求下列不定積分（其中$a$、$b$爲常數）：

（7）$\displaystyle\int\tan^4x\mathrm{d}x;$

解
$\begin{aligned} \displaystyle\int\tan^4x\mathrm{d}x&=\displaystyle\int\tan^2x(\sec^2x-1)\mathrm{d}x\\ &=\displaystyle\int\tan^2x\mathrm{d}(\tan x)-\displaystyle\int(\sec^2x-1)\mathrm{d}x\\ &=\cfrac{1}{3}\tan^3x-\tan x+x+C. \end{aligned}$
（這道題主要利用三角變換公式進行計算）

（10）$\displaystyle\int\sqrt{\cfrac{a+x}{a-x}}\mathrm{d}x;$

解一
$\begin{aligned} \displaystyle\int\sqrt{\cfrac{a+x}{a-x}}\mathrm{d}x&=\displaystyle\int\cfrac{a+x}{\sqrt{a^2-x^2}}\mathrm{d}x=a\displaystyle\int\cfrac{1}{\sqrt{1-\left(\cfrac{x}{a}\right)^2}}\mathrm{d}\left(\cfrac{x}{a}\right)-\cfrac{1}{2}\displaystyle\int\cfrac{\mathrm{d}(a^2-x^2)}{\sqrt{a^2-x^2}}\\ &=a\arcsin\cfrac{x}{a}-\sqrt{a^2-x^2}+C. \end{aligned}$
解二  令$u=\sqrt{\cfrac{a+x}{a-x}}$，即$x=a\cfrac{u^2-1}{u^2+1}$，則
$\begin{aligned} \displaystyle\int\sqrt{\cfrac{a+x}{a-x}}\mathrm{d}x&=\displaystyle\int u\cdot\cfrac{4au}{(1+u^2)^2}\mathrm{d}u=\displaystyle\int-2au\mathrm{d}\left(\cfrac{1}{1+u^2}\right)\\ &=-\cfrac{2au}{1+u^2}+\displaystyle\int\cfrac{2a}{1+u^2}\mathrm{d}u\\ &=-\cfrac{2au}{1+u^2}+2a\arctan u+C\\ &=(x-a)\sqrt{\cfrac{a+x}{a-x}}+2a\arctan\sqrt{\cfrac{a+x}{a-x}}+C. \end{aligned}$
（這道題可以利用換元積分或進行分式變換求解）

（11）$\displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{x(x+1)}};$

解一
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{x(x+1)}}&=\displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{\left(x+\cfrac{1}{2}\right)^2-\left(\cfrac{1}{2}\right)^2}}\\ &\xlongequal{x=-\cfrac{1}{2}+\cfrac{1}{2}\sec u}\displaystyle\int\sec u\mathrm{d}u=\ln|\sec u+\tan u|+C\\ &=\ln|2x+1+2\sqrt{x(x+1)}|+C. \end{aligned}$
解二  當$x>0$時，因爲$\cfrac{1}{\sqrt{x(x+1)}}=\cfrac{1}{x}\sqrt{\cfrac{x}{1+x}}$，故令$u=\sqrt{\cfrac{x}{1+x}}$，即$x=\cfrac{u^2}{1-u^2}$，則
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{x(x+1)}}&=\displaystyle\int\cfrac{2}{1-u^2}\mathrm{d}u=\displaystyle\int\left(\cfrac{1}{1-u}+\cfrac{1}{1+u}\right)\mathrm{d}u\\ &=\ln|\cfrac{1+u}{1-u}|+C=\ln\left|\cfrac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}\right|+C\\ &=\ln|2x+1+2\sqrt{x(x+1)}|+C. \end{aligned}$
  當$x<-1$時，同樣可得$\displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{x(x+1)}}=\ln|2x+1+2\sqrt{x(x+1)}|+C$。（這道題主要用分段或配方進行計算）

（15）$\displaystyle\int\cfrac{\mathrm{d}x}{x^2\sqrt{x^2-1}};$

解  $\displaystyle\int\cfrac{\mathrm{d}x}{x^2\sqrt{x^2-1}}\xlongequal{x=\cfrac{1}{u}}-\displaystyle\int\cfrac{u\mathrm{d}u}{\sqrt{1-u^2}}=\sqrt{1-u^2}+C=\cfrac{\sqrt{x^2-1}}{x}+C,$易知當$x<0$和$x>0$時的結果相同。（這道題主要利用換元積分計算）

（16）$\displaystyle\int\cfrac{\mathrm{d}x}{(a^2-x^2)^{\frac{5}{2}}};$

解  設$x=a\sin u\left(-\cfrac{\pi}{2}<u<\cfrac{\pi}{2}\right)$，則$\sqrt{a^2-x^2}=a\cos u,\mathrm{d}x=a\cos u\mathrm{d}u$，於是
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{(a^2-x^2)^{\frac{5}{2}}}&=\cfrac{1}{a^4}\displaystyle\int\sec^4u\mathrm{d}u=\cfrac{1}{a^4}\displaystyle\int(\tan^2u+1)\mathrm{d}(\tan u)\\ &=\cfrac{\tan^3u}{3a^4}+\cfrac{\tan u}{a^4}+C\\ &=\cfrac{1}{3a^4}\left[\cfrac{x^3}{\sqrt{(a^2-x^2)^3}}+\cfrac{3x}{\sqrt{a^2-x^2}}\right]+C. \end{aligned}$
（這道題主要利用三角函數進行替換）

（17）$\displaystyle\int\cfrac{\mathrm{d}x}{x^4\sqrt{1+x^2}};$

解
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{x^4\sqrt{1+x^2}}&\xlongequal{x=\cfrac{1}{u}}\displaystyle\int\cfrac{-u^3\mathrm{d}u}{\sqrt{1+u^2}}=-\displaystyle\int\left(u\sqrt{1+u^2}-\cfrac{u}{\sqrt{1+u^2}}\right)\mathrm{d}u\\ &=-\cfrac{1}{3}(1+u^2)^{\frac{3}{2}}+\sqrt{1+u^2}+C\\ &=-\cfrac{1}{3}\cfrac{\sqrt{(1+x^2)^3}}{x^3}+\cfrac{\sqrt{1+x^2}}{x}+C. \end{aligned}$
  易知當$x<0$和$x>0$時結果相同。（這道題用倒數換元再拆分因式計算）

（21）$\displaystyle\int\arctan\sqrt{x}\mathrm{d}x;$

解
$\begin{aligned} \displaystyle\int\arctan\sqrt{x}\mathrm{d}x&=\displaystyle\int\arctan\sqrt{x}\mathrm{d}(1+x)=(1+x)\arctan\sqrt{x}-\displaystyle\int\cfrac{1}{2\sqrt{x}}\mathrm{d}x\\ &=(1+x)\arctan\sqrt{x}-\sqrt{x}+C. \end{aligned}$
（這道題在積分時湊出一個因式簡化計算）

（22）$\displaystyle\int\cfrac{\sqrt{1+\cos x}}{\sin x}\mathrm{d}x;$

解
$\begin{aligned} \displaystyle\int\cfrac{\sqrt{1+\cos x}}{\sin x}\mathrm{d}x&=\displaystyle\int\cfrac{\sqrt{2}|\cos\cfrac{x}{2}|}{2\sin\cfrac{x}{2}\cos\cfrac{x}{2}}\mathrm{d}x=\pm\sqrt{2}\displaystyle\int\csc\cfrac{x}{2}\mathrm{d}\left(\cfrac{x}{2}\right)\\ &=\pm\sqrt{2}\ln|\csc\cfrac{x}{2}-\cot\cfrac{x}{2}|+C. \end{aligned}$
  上式當$\cos\cfrac{x}{2}>0$時取正，當$\cos\cfrac{x}{2}<0$時取負。
  當$\cos\cfrac{x}{2}>0$時，
$\ln|\csc\cfrac{x}{2}-\cot\cfrac{x}{2}|=\ln\cfrac{1-\cos\cfrac{x}{2}}{|\sin\cfrac{x}{2}|}\\ =\ln\left(|\csc\cfrac{x}{2}|-|\cot\cfrac{x}{2}|\right).$
  當$\cos\cfrac{x}{2}<0$時，
$\ln|\csc\cfrac{x}{2}-\cot\cfrac{x}{2}|=\ln\cfrac{1-\cos\cfrac{x}{2}}{|\sin\cfrac{x}{2}|}\\ =\ln\left(|\csc\cfrac{x}{2}|+|\cot\cfrac{x}{2}|\right)=-\ln\left(|\csc\cfrac{x}{2}|-|\cot\cfrac{x}{2}|\right).$
  因此有
$\displaystyle\int\cfrac{\sqrt{1+\cos x}}{\sin x}\mathrm{d}x=\sqrt{2}\ln\left(|\csc\cfrac{x}{2}|-|\cot\cfrac{x}{2}|\right)+C.$
（這道題主要利用了分段求解）

（23）$\displaystyle\int\cfrac{x^3}{(1+x^8)^2}\mathrm{d}x;$

解
$\displaystyle\int\cfrac{x^3}{(1+x^8)^2}\mathrm{d}x=\cfrac{1}{4}\displaystyle\int\cfrac{1}{(1+x^8)^2}\mathrm{d}(x^4)\xlongequal{u=x^4}\cfrac{1}{4}\displaystyle\int\cfrac{1}{(1+u^2)^2}\mathrm{d}u.$
  設$u=\tan t\left(-\cfrac{\pi}{2}<t<\cfrac{\pi}{2}\right)$，則$1+u^2=\sec^2t,\mathrm{d}u=\sec^2t\mathrm{d}t$，於是
$\begin{aligned} \text{原式}&=\cfrac{1}{4}\displaystyle\int\cos^2t\mathrm{d}t=\cfrac{2t+\sin2t}{16}+C\\ &=\cfrac{\arctan x^4}{8}+\cfrac{x^4}{8(1+x^8)}+C. \end{aligned}$
（這道題利用了兩次換元求解）

（26）$\displaystyle\int\cfrac{\sin x}{1+\sin x}\mathrm{d}x;$

解一
  令$u=\tan\cfrac{x}{2}$，得
$\begin{aligned} \displaystyle\int\cfrac{\sin x}{1+\sin x}\mathrm{d}x&=\displaystyle\int\cfrac{4u}{(1+u)^2(1+u^2)}\mathrm{d}u=\displaystyle\int\left[\cfrac{-2}{(1+u)^2}+\cfrac{2}{1+u^2}\right]\mathrm{d}u\\ &=\cfrac{2}{1+u}+2\arctan u+C=\cfrac{2}{1+\tan\cfrac{x}{2}}+x+C. \end{aligned}$
解二
$\begin{aligned} \displaystyle\int\cfrac{\sin x}{1+\sin x}\mathrm{d}x&=\displaystyle\int\cfrac{\sin x(1-\sin x)}{\cos^2x}\mathrm{d}x\\ &=-\displaystyle\int\cfrac{1}{\cos^2x}\mathrm{d}(\cos x)-\displaystyle\int(\sec^2x-1)\mathrm{d}x\\ &=\sec x-\tan x+x+C. \end{aligned}$
（這道題主要利用了換元或三角變換求解）

（27）$\displaystyle\int\cfrac{x+\sin x}{1+\cos x}\mathrm{d}x;$

解
$\begin{aligned} \displaystyle\int\cfrac{x+\sin x}{1+\cos x}\mathrm{d}x&=\displaystyle\int\cfrac{x}{2}\sec^2\cfrac{x}{2}\mathrm{d}x+\displaystyle\int\tan\cfrac{x}{2}\mathrm{d}x\\ &=\displaystyle\int x\mathrm{d}(\tan\cfrac{x}{2})+\displaystyle\int\tan\cfrac{x}{2}\mathrm{d}x\\ &=x\tan\cfrac{x}{2}+C. \end{aligned}$
（這道題主要倍角公式求解，另當該函數爲定積分時，有另一種解法，在總習題五第十一題（1），傳送門在這裏）

（28）$\displaystyle\int e^{\sin x}\cfrac{x\cos^3x-\sin x}{\cos^2x}\mathrm{d}x;$

解
$\begin{aligned} \displaystyle\int e^{\sin x}\cfrac{x\cos^3x-\sin x}{\cos^2x}\mathrm{d}x&=\displaystyle\int xe^{\sin x}\cos x\mathrm{d}x-\displaystyle\int e^{\sin x}\tan x\sec x\mathrm{d}x\\ &=\displaystyle\int x\mathrm{d}(e^{\sin x})-\displaystyle\int e^{\sin x}\mathrm{d}(\sec x)\\ &=xe^{\sin x}-\displaystyle\int e^{\sin x}\mathrm{d}x-(\sec xe^{\sin x}-\displaystyle\int e^{\sin x}\mathrm{d}x)\\ &=(x-\sec x)e^{\sin x}+C. \end{aligned}$
（這道題利用分部積分法化簡至有相同部分可以相互抵消求解）

（31）$\displaystyle\int\cfrac{e^{3x}+e^x}{e^{4x}-e^{2x}+1}\mathrm{d}x;$

解
$\begin{aligned} \displaystyle\int\cfrac{e^{3x}+e^x}{e^{4x}-e^{2x}+1}\mathrm{d}x&=\displaystyle\int\cfrac{e^{x}+e^{-x}}{e^{2x}+e^{-2x}-1}\mathrm{d}x=\displaystyle\int\cfrac{\mathrm{d}(e^{x}-e^{-x})}{(e^{x}-e^{-x})^2+1}\\ &=\arctan(e^{x}-e^{-x})+C. \end{aligned}$
（這道題利用分式約分求解）

（33）$\displaystyle\int\ln^2(x+\sqrt{1+x^2})\mathrm{d}x;$

解
$\begin{aligned} \displaystyle\int\ln^2(x+\sqrt{1+x^2})\mathrm{d}x&=x\ln^2(x+\sqrt{1+x^2})-\displaystyle\int\cfrac{2x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm{d}x\\ &=x\ln^2(x+\sqrt{1+x^2})-\displaystyle\int2\ln(x+\sqrt{1+x^2})\mathrm{d}(\sqrt{1+x^2})\\ &=x\ln^2(x+\sqrt{1+x^2})-2\sqrt{1+x^2}\ln(x+\sqrt{1+x^2})+2x+C. \end{aligned}$
（這道題主要利用$(\ln(x+\sqrt{1+x^2}))'=\cfrac{1}{\sqrt{1+x^2}}$進行化簡求解）

（34）$\displaystyle\int\cfrac{\ln x}{(1+x^2)^{\frac{3}{2}}}\mathrm{d}x;$

解
$\begin{aligned} \displaystyle\int\cfrac{\ln x}{(1+x^2)^{\frac{3}{2}}}&\xlongequal{x=\cfrac{1}{u}}\displaystyle\int\cfrac{u\ln u}{(1+u^2)^{\frac{3}{2}}}\mathrm{d}u=-\displaystyle\int\ln u\mathrm{d}((1+u^2)^{-\frac{1}{2}})\\ &=-\cfrac{\ln u}{\sqrt{1+u^2}}+\displaystyle\int\cfrac{\mathrm{d}u}{u\sqrt{1+u^2}}\\ &=x\cfrac{\ln x}{\sqrt{1+x^2}}-\displaystyle\int\cfrac{\mathrm{d}x}{\sqrt{1+x^2}}\\ &=x\cfrac{\ln x}{\sqrt{1+x^2}}-\ln(x+\sqrt{1+x^2})+C. \end{aligned}$
（這道題利用倒數換元求解）

（35）$\displaystyle\int\sqrt{1-x^2}\arcsin x\mathrm{d}x;$

解  設$x=\sin u\left(-\cfrac{\pi}{2}<u<\cfrac{\pi}{2}\right)$，則$\sqrt{1-x^2}=\cos u,\mathrm{d}x=\cos u\mathrm{d}u$，於是
$\begin{aligned} \displaystyle\int\sqrt{1-x^2}\arcsin x\mathrm{d}x&=\displaystyle\int u\cos^2u\mathrm{d}u=\cfrac{1}{2}\displaystyle\int u(1+\cos2u)\mathrm{d}u\\ &=\cfrac{1}{4}\displaystyle\int u\mathrm{d}(2u+\sin2u)\\ &=\cfrac{u(2u+\sin2u)}{4}-\cfrac{1}{4}\displaystyle\int(2u+\sin2u)\mathrm{d}u\\ &=\cfrac{u^2}{4}+\cfrac{u}{4}\sin2u-\cfrac{\sin^2u}{4}+C\\ &=\cfrac{(\arcsin x)^2}{4}+\cfrac{x}{2}\sqrt{1-x^2}\arcsin x-\cfrac{x^2}{4}+C. \end{aligned}$
（這道題利用三角函數換元求解）

（36）$\displaystyle\int\cfrac{x^3\arccos x}{\sqrt{1-x^2}}\mathrm{d}x;$

解  設$x=\cos u\left(0<u<\pi\right)$，則$\sqrt{1-x^2}=\sin u,\mathrm{d}x=-\sin u\mathrm{d}u$，於是
$\begin{aligned} \displaystyle\int\cfrac{x^3\arccos x}{\sqrt{1-x^2}}\mathrm{d}x&=-\displaystyle\int u\cos^3u\mathrm{d}u=-\displaystyle\int u\mathrm{d}(\sin u-\cfrac{1}{3}\sin^3u)\\ &=-u(\sin u-\cfrac{1}{3}\sin^3u)+\displaystyle\int (\sin u-\cfrac{1}{3}\sin^3u)\mathrm{d}u\\ &=-u(\sin u-\cfrac{1}{3}\sin^3u)-\cfrac{1}{3}\displaystyle\int(2+\cos^2u)\mathrm(\cos u)\\ &=-u(\sin u-\cfrac{1}{3}\sin^3u)-\cfrac{2}{3}\cos u-\cfrac{1}{9}\cos^3u+C\\ &=-\cfrac{1}{3}\sqrt{1-x^2}(2+x^2)\arccos x-\cfrac{1}{9}(6+x^2)+C. \end{aligned}$
（這道題主要用三角函數換元再分部積分求解）

（38）$\displaystyle\int\cfrac{\mathrm{d}x}{\sin^3x\cos x};$

解
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{\sin^3x\cos x}&=-\displaystyle\int\cot x\sec^2x\mathrm{d}(\cot x)\xlongequal{u=\cot x}-\displaystyle\int u\left(1+\cfrac{1}{u^2}\right)\mathrm{d}u\\ &=-\cfrac{u^2}{2}-\ln|u|+C=-\cfrac{\cot^2x}{2}-\ln|\cot x|+C. \end{aligned}$
（這道題主要利用了三角函數換元）

（39）$\displaystyle\int\cfrac{\mathrm{d}x}{(2+\cos x)\sin x};$

解
$\begin{aligned} \displaystyle\int\cfrac{\mathrm{d}x}{(2+\cos x)\sin x}&=\displaystyle\int\cfrac{\mathrm{d}(\cos x)}{(2+\cos x)(\cos^2x+1)}\\ &\xlongequal{u=\cos x}\displaystyle\int\cfrac{\mathrm{d}u}{(2+u)(u^2+1)}\\ &=\displaystyle\int\left[\cfrac{1}{6(u-1)}-\cfrac{1}{2(u+1)}+\cfrac{1}{3(u+2)}\right]\mathrm{d}u\\ &=\cfrac{1}{6}\ln|u-1|-\cfrac{1}{2}\ln|u+1|+\cfrac{1}{3}\ln|u+2|+C\\ &=\cfrac{1}{6}\ln(1-\cos x)-\cfrac{1}{2}\ln(1+\cos x)+\cfrac{1}{3}\ln(2+\cos x)+C. \end{aligned}$
（這道題主要利用三角函數換元和因式分解積分）

（40）$\displaystyle\int\cfrac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x;$

解一
$\begin{aligned} \displaystyle\int\cfrac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x&=\displaystyle\int\cfrac{\cfrac{1}{2}(\sin x+\cos x)^2-\cfrac{1}{2}}{\sin x+\cos x}\mathrm{d}x\\ &=\cfrac{1}{2}\displaystyle\int(\sin x+\cos x)\mathrm{d}x-\cfrac{1}{2}\displaystyle\int\cfrac{1}{\sin x+\cos x}\mathrm{d}x\\ &=\cfrac{1}{2}(-\cos x+\sin x)-\cfrac{1}{2}\displaystyle\int\cfrac{1}{\sin x+\cos x}\mathrm{d}x. \end{aligned}$
  令$u=\tan\cfrac{x}{2}$，則$\sin x=\cfrac{2u}{1+u^2},\cos x=\cfrac{1-u^2}{1+u^2},\mathrm{d}x=\cfrac{2}{1+u^2}\mathrm{d}u$，故有
$\begin{aligned} \displaystyle\int\cfrac{1}{\sin x+\cos x}\mathrm{d}x&=\displaystyle\int\cfrac{2}{2u+1-u^2}\mathrm{d}u=-\displaystyle\int\cfrac{2}{(u-1)^2-(\sqrt{2})^2}\mathrm{d}u\\ &=-\cfrac{1}{\sqrt{2}}\displaystyle\int\cfrac{2}{u-1-\sqrt{2}}\mathrm{d}u+\cfrac{1}{\sqrt{2}}\displaystyle\int\cfrac{2}{u-1+\sqrt{2}}\mathrm{d}u\\ &=\cfrac{1}{\sqrt{2}}\ln\left|\cfrac{u-1+\sqrt{2}}{u-1-\sqrt{2}}\right|+C'. \end{aligned}$
  因此有
$\displaystyle\int\cfrac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x=\cfrac{1}{2}(\sin x-\cos x)-\cfrac{1}{2\sqrt{2}}\ln\left|\cfrac{\tan\cfrac{x}{2}-1+\sqrt{2}}{\tan\cfrac{x}{2}-1-\sqrt{2}}\right|+C.$
解二
$\begin{aligned} \displaystyle\int\cfrac{\sin x\cos x}{\sin x+\cos x}\mathrm{d}x&=\displaystyle\int\cfrac{\sin x\cos x}{\sqrt{2}\sin\left(x+\cfrac{\pi}{4}\right)}\mathrm{d}x\xlongequal{u=x+\cfrac{\pi}{4}}\displaystyle\int\cfrac{2\sin^2u-1}{2\sqrt{2}\sin u}\mathrm{d}u\\ &=\cfrac{1}{\sqrt{2}}\displaystyle\int\sin u\mathrm{d}u-\cfrac{1}{2\sqrt{2}}\displaystyle\int\csc u\mathrm{d}u\\ &=-\cfrac{\cos\left(x+\cfrac{\pi}{4}\right)}{\sqrt{2}}-\cfrac{1}{2\sqrt{2}}\ln\left|\csc\left(x+\cfrac{\pi}{4}\right)-\cot\left(x+\cfrac{\pi}{4}\right)\right|+V. \end{aligned}$
（這道題主要利用萬能公式或三角函數換元積分求解）

寫在最後

  如果覺得文章不錯就點個贊吧。另外，如果有不同的觀點，歡迎留言或私信。
   歡迎非商業轉載，轉載請註明出處。
  另，參考的積分表及公式見附錄。