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标题: 两个相等的定积分 [打印本页]

作者: 青青子衿    时间: 2018-5-23 17:01     标题: 两个相等的定积分

证明:\[\int_0^{2\pi}(1+\sin\theta\cos\theta)^{\sigma-1}\rmd{\theta}=\int_0^{2\pi}\dfrac{\dfrac{2\sqrt{3}}{3}}{\left(\dfrac{2}{3}\cos^2t+2\sin^2t\right)^{\sigma}}\rmd{t}\]
作者: abababa    时间: 2018-5-28 18:22

本帖最后由 abababa 于 2018-5-28 18:24 编辑

回复 1# 青青子衿

不太感兴趣这个,发网友的解答看看:
\begin{align*}
\int_{0}^{2\pi}(1+\sin x\cos x)^{\sigma-1}dx &= \frac{1}{2^{\sigma-1} \cdot 2}\int_{0}^{2\pi}(2+\sin 2x)^{\sigma-1}d2x = \frac{1}{2^{\sigma}}\int_{0}^{4\pi}(2+\sin x)^{\sigma-1}dx\\
&= \frac{1}{2^{\sigma}} \cdot 2\int_{-\pi}^{\pi}(2+\sin x)^{\sigma-1}dx = \frac{1}{2^{\sigma}} \cdot 2\int_{-\pi}^{\pi}(2+\cos x)^{\sigma-1}dx\\
&=->(x=2\arctan s)= \frac{1}{2^{\sigma}} \cdot 2\int_{-\infty}^{+\infty}(\frac{s^2+3}{s^2+1})^{\sigma-1} \cdot \frac{2}{s^2+1}ds\\
&= ->(s=\sqrt{3}s)= \frac{1}{2^{\sigma}}\cdot 2\cdot(\sqrt{3})^{2\sigma-1}\int_{-\infty}^{+\infty}(\frac{1+s^2}{1+3s^2})^{\sigma-1}\cdot\frac{2}{3s^2+1}ds
\end{align*}


\begin{align*}
\int_{0}^{2\pi}\frac{\frac{2}{\sqrt{3}}}{(\frac{2}{3}\cos^2x+2\sin^2x)^{\sigma}}dx &= \frac{2}{\sqrt{3}} \cdot \frac{1}{2}\int_{0}^{2\pi}(\frac{2}{3})^{-\sigma}(2-\cos 2x)^{-\sigma}d2x = \frac{(\sqrt{3})^{2\sigma-1}}{2^{\sigma}} \cdot 2\int_{-\pi}^{\pi}(2-\cos x)^{-\sigma}dx\\
&=->(x=2\arctan s)=\frac{(\sqrt{3})^{2\sigma-1}}{2^{\sigma}} \cdot 2 \int_{-\infty}^{\infty}(\frac{s^2+1}{3 s^2+1})^{\sigma}\cdot \frac{2}{s^2+1}ds
\end{align*}




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