求定积分 ∫(-π,π)xsinx[arctan(e^x)+∫(0,x)e^(t^2)dt]／[1+(cosx)^2]dx

马奕琛

\(试求\displaystyle \int_{-π}^{π}\dfrac {xsinx(arctane^{x}+\int_{0}^{x}e^{t^{2}}dt)}{1+cos^{2}x}dx\)

Future_maths

表面上看起来很吓人的题目，实际上很简单。

fungarwai

\(\displaystyle \tan(\frac{\pi}{2}-\theta)=\frac{1}{\tan \theta}\)

\(\displaystyle \arctan x+\arctan \frac{1}{x}=\frac{\pi}{2}\)

\(\displaystyle \int_{-\pi}^{\pi}\dfrac {x\sin x(\arctan e^{x}+\int_{0}^{x}e^{t^{2}}dt)}{1+\cos^{2}x}dx\)
\(\displaystyle =\int_{0}^{\pi}\dfrac {x\sin x(\arctan e^{x}+\int_{0}^{x}e^{t^{2}}dt)}{1+\cos^{2}x}dx
+\int_{0}^{\pi}\dfrac {(-x)\sin(-x)(\arctan e^{-x}+\int_{0}^{-x}e^{t^{2}}dt)}{1+\cos^{2}(-x)}dx\)
\(\displaystyle =\int_{0}^{\pi}\dfrac {x\sin x(\arctan e^{x}+\arctan e^{-x})}{1+\cos^{2}x}dx\)
\(\displaystyle =\frac{\pi}{2}\int_{0}^{\pi}\dfrac {x\sin x}{1+\cos^{2}x}dx\)

\(\displaystyle =\frac{\pi}{4}\int_{0}^{\pi}\dfrac {x\sin x}{1+\cos^{2}x}dx
+\frac{\pi}{4}\int_{0}^{\pi}\dfrac {(\pi-x)\sin(\pi-x)}{1+\cos^{2}(\pi-x)}dx\)
\(\displaystyle =\frac{\pi^2}{4}\int_{0}^{\pi}\dfrac {\sin x}{1+\cos^{2}x}dx\)
\(\displaystyle \left.=\frac{-\pi^2}{4}\arctan(\cos x)\right|_{0}^{\pi}\)
\(\displaystyle =\frac{\pi^3}{8}\)

luyuanhong

楼上 fungarwai 的解答很好！已收藏。