笔记下载：三角函数的积分
三角函数的积分
1，\(\displaystyle\int \sin ^m xdx\)，\(\displaystyle\int \cos^m xdx\)
（1）m 为奇数\(\quad\Rightarrow\quad\)换元法，首先提出一次方，然后利用 \(\displaystyle\sin^2 x+\cos^2 x=1\)
（2）m 为偶数\(\quad\Rightarrow\quad\)倍角公式降阶，\(\displaystyle\sin^2 x=\frac{1-\cos 2x}{2}\)，\(\displaystyle\cos^2 x=\frac{1+\cos 2x}{2}\)
例1，求\(\displaystyle\int \sin^3 xdx\)
解答：\(\displaystyle\int \sin^3 xdx\)
\(\displaystyle\quad\ =\int \sin^2 x\sin x dx\)
\(\displaystyle\quad\ =\int (1-\cos^2 x) \sin x dx\qquad\qquad\)
\(\qquad\qquad\)我们令 \(\displaystyle u=\cos x\)，\(\displaystyle du=-\sin x dx\)，则
\(\displaystyle\quad\ =\int (1-u^2)(-du)\)
\(\displaystyle\quad\ =\frac{1}{3}u^3-u+C\)
\(\displaystyle\quad\ =\frac{1}{3} \cos^3 x-\cos x+C\)
例2，求 \(\displaystyle\int \cos^2 xdx\)
解答：\(\displaystyle\int \cos^2 xdx\)
\(\displaystyle\quad\ =\int \frac{1+\cos 2x}{2} dx\)
\(\displaystyle\quad\ =\frac{1}{2}x+\frac{1}{4}\sin 2x+C\)
注：\(\displaystyle\int \sin^m xdx\)、\(\displaystyle\int \cos^m xdx\) 也可以用递推法
2，\(\displaystyle\int \sin^m x \cos^n xdx\)
（1）m 或 n 有一个为奇数，提出一个一次方，然后换元
（2）m 和 n 都是偶数，降阶
（3）m=n 为偶数，\(\displaystyle\sin x\cos x=\frac{1}{2}\sin 2x\)
例3，求\(\displaystyle\int \sin^2 x \cos^3 x dx\)
解答：\(\displaystyle\int \sin^2 x \cos^3 x dx\)
\(\displaystyle\quad\ =\int \sin^2 x\cos^2 x\cdot\cos xdx\)
\(\displaystyle\quad\ =\int \sin^2 x (1-\sin^2 x)\cdot\cos xdx\)
\(\qquad\qquad\)我们令 \(\displaystyle u=\sin x\)，\(\displaystyle du=\cos xdx\)，则
\(\displaystyle\quad\ =\int u^2 (1-u^2) du\)
\(\displaystyle\quad\ =\frac{1}{3} u^3-\frac{1}{5} u^5+C\)
\(\displaystyle\quad\ =\frac{1}{3}\sin^3 x-\frac{1}{5}\sin^5 x +C\)
例4，求 \(\displaystyle\int \sin^2 x \cos^4 xdx\)
解答：\(\displaystyle\int \sin^2 x \cos^4 xdx\)
\(\displaystyle\quad\ =\int \frac{1-\cos 2x}{2}\cdot\left(\frac{1+\cos 2x}{2}\right)^2 dx\)
\(\displaystyle\quad\ =\frac{1}{8}\int (1-\cos 2x)(1+2\cos2x+\cos^2 2x)dx\)
\(\displaystyle\quad\ =\frac{1}{8}\int (1+\cos2x-\cos^2 2x-\cos^3 2x)dx\)
\(\displaystyle\quad\ =\frac{1}{8}x+\frac{1}{16}\sin 2x-\int \cos^2 2xdx-\int \cos^3 2xdx\)
\(\displaystyle\quad\ =\frac{1}{8}x+\frac{1}{16}\sin 2x-\int \frac{1+\cos 4x}{2}dx-\int \cos^2 2x\cdot\cos 2xdx\)
\(\displaystyle\quad\ =\frac{1}{8}x+\frac{1}{16}\sin 2x-\frac{1}{2}x-\frac{1}{8}\sin 4x-\int (1-\sin^2 2x)\cos 2xdx\)
\(\qquad\qquad\)我们令 \(\displaystyle u=\sin 2x\)，\(\displaystyle du=2\cos 2xdx\)，则
\(\displaystyle\quad\ =-\frac{3}{8}x+\frac{1}{16}\sin 2x-\frac{1}{8}\sin 4x-\int \frac{1}{2}(1-u^2)du\)
\(\displaystyle\quad\ =-\frac{3}{8}x+\frac{1}{16}\sin 2x-\frac{1}{8}\sin 4x-\frac{1}{2}x+\frac{1}{6}u^3+C\)
\(\displaystyle\quad\ =-\frac{3}{8}x+\frac{1}{16}\sin 2x-\frac{1}{8}\sin 4x-\frac{1}{2}x+\frac{1}{6}\sin^3 2x+C\)
\(\displaystyle\quad\ =-\frac{7}{8}x+\frac{1}{16}\sin 2x+\frac{1}{6}\sin^3 2x-\frac{1}{8}\sin 4x+C\)
例5，求\(\displaystyle\int\sin^2x\cos^2xdx\)
解答：\(\displaystyle\int\sin^2x\cos^2xdx\)
\(\displaystyle\quad\ =\int \left(\sin x\cos x\right)^2dx\)
\(\displaystyle\quad\ =\int \left(\frac{1}{2}\sin 2x\right)^2 dx\)
\(\displaystyle\quad\ =\frac{1}{4}\int\sin^2 2xdx\)
\(\displaystyle\quad\ =\frac{1}{4}\int \frac{1-\cos4x}{2}dx\)
3，\(\displaystyle\int tan^m x\sec^n xdx\)
（1）n 是偶数，分一个\(\displaystyle\sec^2 x\) 出来 \(\displaystyle\quad\Rightarrow\quad\)换元
（2）m 是奇数，分一个\(\displaystyle \tan x\cdot\sec x\) 出来 \(\displaystyle\quad\Rightarrow\quad\)换元
例6，求\(\displaystyle\int \tan^2x\cdot\sec^4 xdx\)
解答：\(\displaystyle\int \tan^2x\cdot\sec^4 xdx\)
\(\displaystyle\quad\ =\int \tan^2x\sec^2x\cdot\sec^2x dx\)
\(\displaystyle\quad\ =\int \tan^2 x\cdot\left(\tan^2+1\right) \sec^2 xdx\)
\(\qquad\qquad\)我们令 \(\displaystyle u=\tan x\)，\(\displaystyle du=\sec^2 xdx\)，则
\(\displaystyle\quad\ =\int u^2(u^2+1)du\)
\(\displaystyle\quad\ =\frac{1}{5}u^5+\frac{1}{3}u^3+C\)
\(\displaystyle\quad\ =\frac{1}{5}\tan^5 x+\frac{1}{3}\tan^x+C\)
例7，求\(\displaystyle\int \tan^3x \sec x dx\)
解答：\(\displaystyle\int \tan^3x \sec x dx\)
\(\displaystyle\quad\ =\int \tan^2 x\left(\tan x\cdot\sec x\right)dx\)
\(\qquad\qquad\)我们令 \(\displaystyle u=\sec x\)，\(\displaystyle du=\sec x\tan x dx\)，则
\(\displaystyle\quad\ =\int (u^2-1)du\)
\(\displaystyle\quad\ =\frac{1}{3}u^3-u+C\)
\(\displaystyle\quad\ =\frac{1}{3}\sec^3x-\sec^x+C\)
4，积化和差
公式回顾：\(\displaystyle\sin (A+B)=\sin A\cos B+\cos A\sin B\)
\(\displaystyle\qquad\qquad\ \ \sin (A-B)=\sin A\cos B-\cos A\sin B\)
\(\displaystyle\qquad\qquad\ \ \cos (A+B)=\cos A\cos B-\sin A\sin B\)
\(\displaystyle\qquad\qquad\ \ \cos (A-B)=\cos A\cos B+\sin A\sin B\)
\(\displaystyle\qquad\qquad\ \ \sin A\sin B=\frac{\cos (A-B)-\cos (A+B)}{2}\)
\(\displaystyle\qquad\qquad\ \ \cos A\cos B=\frac{\cos (A-B)+\cos (A+B)}{2}\)
例8，求\(\displaystyle\int\sin 4x \sin 5x dx\)
解答：\(\displaystyle\int\sin 4x \sin 5x dx\)
\(\displaystyle\quad\ =\int \frac{1}{2}\left[\cos (-x)-\cos 9x\right]dx\)
\(\displaystyle\quad\ =\frac{1}{2}\left(-\sin(-x)-\frac{1}{9}\sin 9x\right)+C\)
\(\displaystyle\quad\ =\frac{1}{2}\sin x-\frac{1}{18}\sin 9x+C\)
5，半角代换
公式回顾：\(\displaystyle t=\tan \frac{x}{2}\)，\(\displaystyle \sin x=\frac{2t}{1+t^2}\)，\(\displaystyle \cos x=\frac{1-t^2}{1+t^2}\)
\(\displaystyle\qquad\qquad\ \ dx=\frac{2dt}{1+t^2}\)，\(\displaystyle\frac{x}{2}=\arctan t\)，\(\displaystyle x=2\arctan t\)
例9，求\(\displaystyle\int \csc xdx\)
解答：\(\displaystyle\int \csc xdx\)
\(\displaystyle\quad\ =\int \frac{1}{\sin x}dx\)
\(\displaystyle\quad\ =\int \frac{1+t^2}{2t}\cdot\frac{2dt}{1+t^2}\)
\(\displaystyle\quad\ =\int \frac{1}{t}dt\)
\(\displaystyle\quad\ =\ln|t|+C\)
\(\displaystyle\quad\ =\ln|\tan \frac{x}{2}|+C\)
\(\qquad\qquad\)而 \(\displaystyle \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{2\sin^2 \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac {1-\cos x}{\sin x}=\csc x-\cot x\)，则
\(\displaystyle\int \csc xdx=\ln|\csc x-\cot x|+C\)
6，求\(\displaystyle \frac{1}{\csc x}dx\)
解答：\(\displaystyle \frac{1}{\csc x}dx\)
\(\displaystyle\quad =\int \frac{dx}{\sin x}\)
\(\displaystyle\quad =\int \frac{\sin x}{\sin^2 x}dx\)
\(\displaystyle\quad =\int \frac {\sin x}{1-\cos ^2 x}dx\)
\(\qquad\qquad\)我们令 \(\displaystyle u=\cos x\)，\(\displaystyle du=-\sin x dx\)，则
\(\displaystyle\quad=\int \frac{du}{u^2-1}\)
注：参见 菲赫金哥尔茨 著《微积分学教程》中有理函数积分方法