三角函数的积分

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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