微积分(A)随缘一题[30]

计算：\(\int_0^\pi \sin^6x\cos^2xdx\)

首先有：

\[\int \sin^6x\cos^2 x dx=\int (\sin^6x-\sin^8x)dx \\ \]

其次：

\[\begin{aligned} I_n =&\int \sin^ndx \\ =&-\int \sin^{n-1}d\cos x \\ =&-\sin^{n-1}x\cos x+\int \cos x(n-1)\sin^{n-2}\cos xdx \\ =&-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}(1-\sin^2 x)dx \\ =&-\sin^{n-1}x\cos x+(n-1)I_{n-2}-(n-1)I_{n} \\ I_n=&\frac{n-1}{n}I_{n-2}-\frac{1}{n}\sin^{n-1}x\cos x+C \end{aligned} \]

化简得：

\[\int (\sin^6x-\sin^8x)dx=I_6-I_8=I_6-(\frac{7}{8}I_6-\frac{\sin^7x\cos x}{8})=\frac{1}{8}I_6+\frac{\sin^7x\cos x}{8} \]

同时有：

\[\begin{aligned} &I_0=x+C_0 \\ &I_2=\frac{1}{2}I_0-\frac{\sin x\cos x}{2}=\frac{x-\sin x\cos x}{2}+C_2 \\ &I_4=\frac{3}{4}I_2-\frac{\sin^3x\cos x}{4}=\frac{3}{8}x-\frac{3}{8}\sin x\cos x-\frac{\sin^3x \cos x}{4}+C_4 \\ &I_6=\frac{5}{6}I_4-\frac{\sin^5x\cos x}{6}=\frac{5x}{16}-\frac{5(\sin x\cos x)}{16}-\frac{5(\sin^3x\cos x)}{24}-\frac{\sin^5 x\cos x}{6}+C_6 \end{aligned} \]

代入原式：

\[\begin{aligned} &\int _0^{\pi}\sin^6x\cos^2xdx=\left(\frac{1}{8}I_6+\frac{\sin^7x\cos x}{8}\right) \bigg|_0^{\pi} \\ = & \left( \frac{5}{128}x-\frac{5}{128}\sin x\cos x-\frac{5}{192}\sin ^3x\cos x-\frac{1}{48}\sin^5x\cos x+\frac{1}{8}\sin^7x\cos x \right)\bigg|_0^{\pi} \\ =&\frac{5\pi}{128} \end{aligned} \]

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实际上：

\[I_n \bigg|_0^\pi=\left(\frac{n-1}{n}I_{n-2}-\frac{1}{n}\sin^{n-1}x\cos x+C\right)\bigg|_0^\pi = \begin{cases} \pi & \quad (n=0)\\ 2 & \quad(n=1)\\ \frac{n-1}{n}I_{n-2} & \quad (n \ge 2) \end{cases} \]

所以：

\[\begin{cases} I_{2n} \bigg|_0^{\pi}=\pi \cdot \frac{(2n-1)!!}{(2n)!!} \\ I_{2n+1}\bigg|_0^{\pi}=2 \cdot \frac{(2n)!!}{(2n+1)!!} \end{cases} \]

所以：

\[\begin{aligned} &\int _0^{\pi}\sin^6x\cos^2xdx=\left(\frac{1}{8}I_6+\frac{\sin^7x\cos x}{8}\right) \bigg|_0^{\pi} \\ =&\frac{1}{8}\pi \cdot \frac{5!!}{6!!} & \\ =&\frac{5\pi}{128} \end{aligned} \]

实际上就是wallis公式？