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如何求 $$\sin^n x, \cos^n x$$ 的积分？

$$\sin^n x$$ 和 $$\cos^n x$$ 的积分方法主要有两种：第一种是根据 $$n$$ 的奇、偶情况分别采用换元法或者降阶法来求；另一种是递推法。

\begin{align*}\int\sin^5xdx&=\int\sin^4x\sin xdx\\ &=\int(1-\cos^2x)^2(-\cos x)’dx\\ &=-\int(1-2\cos^2x+\cos^4x)d(\cos x)\\ &=-\int(1-2u^2+u^4)du\\ &=-(u-\frac{2}{3}u^3+\frac{1}{u^5})+C\\ &= \frac{2}{3} \cos^3x-\cos x-\frac{1}{5}\cos^5x+C\end{align*}

$\int\sin^nxdx=\int\left( \frac{1-\cos(2x)}{2} \right)^{n/2}dx$

$\int\cos^nxdx=\int\left( \frac{1+\cos(2x)}{2} \right)^{n/2}dx$

\begin{align*}\int\cos^4xdx&=\int \left( \frac{1+\cos(2x)}{2} \right)^2dx \\ &=\frac{1}{4}\int\left(1+2\cos(2x)+\cos^2(2x)\right)dx\\ &= \frac{1}{4}\int\left(1+2\cos(2x)+\frac{1+\cos(4x)}{2}\right)dx\\ &= \frac{1}{4}\left(x+\sin(2x)+\frac{x}{2}+\frac{1}{8}\sin(4x)\right)+C \end{align*}

\begin{align*}\int\sin^nxdx&=\int\sin^{n-1}x\sin xdx\\ &=-\sin^{n-1}x\cos x+\int(n-1)\sin^{n-2}x\cos x\cos xdx\\ & = -\sin^{n-1}x\cos x+ (n-1)\int\sin^{n-2}x\cos^2xdx\\ &= -\sin^{n-1}x\cos x+ (n-1)\int \sin^{n-2}x(1-\sin^2x)dx\\ &= -\sin^{n-1}x\cos x+ (n-1)\int (\sin^{n-2}x-\sin^nx)dx \end{align*}

$n\int\sin^nxdx= -\sin^{n-1}x\cos x+ (n-1)\int \sin^{n-2}xdx$

$I_n= -\frac{1}{n}\sin^{n-1}x\cos x+\frac{n-1}{n}I_{n-2}$

$\int\sin^0xdx=x+C, \quad \int\sin xdx=-\cos x+C$

\begin{align*}\int\sin^5xdx&= -\frac{1}{5}\sin^{4}x\cos x+\frac{4}{5}\int\sin^3xdx\\ &= -\frac{1}{5}\sin^{4}x\cos x +\frac{4}{5} \left(-\frac{1}{3}\sin^2x\cos x+\frac{2}{3}\int\sin xdx\right)\\ &= -\frac{1}{5}\sin^{4}x\cos x -\frac{4}{15} \sin^2x\cos x -\frac{8}{15}\cos x+C \end{align*}

$$\int\cos^nxdx$$ 可以完全同样的方式处理。我们有

\begin{align*}\int\cos^n xdx&=\int\cos^{n-1}x\cos xdx\\ &=\cos^{n-1}x\sin x+\int(n-1)\cos^{n-2}x\sin^2xdx\\ &= \cos^{n-1}x\sin x+ (n-1) \int\cos^{n-2}x(1-\cos^2x)dx \\ &= \cos^{n-1}x\sin x+ (n-1) \int\cos^{n-2}xdx- (n-1) \int\cos^nxdx\end{align*}

$\int\cos^nxdx=\frac{1}{n} \cos^{n-1}x\sin x + \frac{n-1}{n} \int\cos^{n-2}xdx$

\begin{align*}\int\cos^4xdx&=\frac{1}{4}\cos^3x\sin x+ \frac{2}{3} \int\cos^2xdx\\ &= \frac{1}{4}\cos^3x\sin x+ \frac{2}{3} \left(\frac{1}{2}\cos x\sin x+\frac{1}{2}\int\cos^0xdx\right)\\ &= \frac{1}{4}\cos^3x\sin x+ \frac{1}{3} \cos x\sin x +\frac{1}{3}x+C \end{align*}