一例积分题

求积分：

雨中漫步

$\int_0^{\frac{\pi}{2}}{\frac{x\sin x}{1+\cos ^2x}dx}$

解：首先通过分部积分：
$\int_0^{\frac{\pi}{2}}{\frac{x\sin x}{1+\cos ^2x}dx}$
$\qquad=-\int_0^{\frac{\pi}{2}}{xd\arctan\cos x}$
$\quad =\int_0^{\frac{\pi}{2}}{\arctan\cos xdx}$
$=\int_0^1{\frac{\arctan x}{\sqrt{1-x^2}}dx}$

设$I\left( \alpha \right) =\int_0^1{\frac{\arctan \alpha x}{\sqrt{1-x^2}}dx},$我们有：
$I\left( 0 \right) =0,$
$I\left( \infty \right) =\frac{\pi}{2}\int_0^1{\frac{dx}{\sqrt{1-x^2}}}=\frac{\pi ^2}{4}$
接着我们研究导数有：
$I'\left( \alpha \right) =\int_0^1{\frac{x}{\left( 1+\alpha ^2x^2 \right) \sqrt{1-x^2}}dx}$
$=\int_0^1{\frac{dt}{1+\alpha ^2-\alpha ^2t^2}}$
$=\frac{\ln \left( \alpha +\sqrt{\alpha ^2+1} \right)}{\alpha \sqrt{1+\alpha ^2}}$
那么：
$\int_0^1{\frac{\arctan x}{\sqrt{1-x^2}}dx}=I\left( 1 \right) =\int_0^1{I'\left( \alpha \right)}d\alpha \\ =\int_0^1{\frac{\ln \left( \alpha +\sqrt{\alpha ^2+1} \right)}{\alpha \sqrt{1+\alpha ^2}}}d\alpha \\ =-\int_0^1{\frac{\ln \left( \alpha +\sqrt{\alpha ^2+1} \right)}{\sqrt{1+\frac{1}{\alpha ^2}}}}d\frac{1}{\alpha}\\ =-\int_0^1{\ln \left( \alpha +\sqrt{\alpha ^2+1} \right) d\ln \left( \frac{1}{\alpha}+\sqrt{\left( \frac{1}{\alpha} \right) ^2+1} \right)}\\ =\int_0^1{\frac{\ln \left( \frac{1}{\alpha}+\sqrt{\left( \frac{1}{\alpha} \right) ^2+1} \right)}{\sqrt{\alpha ^2+1}}d\alpha -\ln ^2\left( 1+\sqrt{2} \right)}\\ =\int_1^{+\infty}{\frac{\ln \left( \alpha +\sqrt{\alpha ^2+1} \right)}{\alpha \sqrt{1+\alpha ^2}}d\alpha -\ln ^2\left( 1+\sqrt{2} \right)}\\ =\int_1^{+\infty}{I'\left( \alpha \right) d}\alpha -\ln ^2\left( 1+\sqrt{2} \right) \\ =I\left( \infty \right) -I\left( 1 \right) -\ln ^2\left( 1+\sqrt{2} \right) =I\left( 1 \right) \\ =\frac{I\left( \infty \right) -\ln ^2\left( 1+\sqrt{2} \right)}{2}\\ =\frac{\pi ^2}{8}-\frac{1}{2}\ln ^2\left( 1+\sqrt{2} \right)$

来源：CSDN

作者：abcwsp

链接：https://blog.csdn.net/abcwsp/article/details/104147436