Mathematics

# Solve $I=\int_{-\pi}^{\pi}{\dfrac{2x(\sin x+1)}{1+\cos^{2}x}dx}$

##### SOLUTION
$I=\displaystyle \int _{-\pi/2}^{\pi}= \dfrac{2x\left(\sin x+1\right)}{1+\cos^{2}x}dx$
$=\displaystyle \int _{0}^{\pi}= \dfrac{2x\left(1+\sin x\right)}{1+\cos^{2}x}dx + \displaystyle \int _{0}^{\pi}= \dfrac{-2x\left(1-\sin x\right)}{1+\cos^{2}x}dx$
$I=\displaystyle \int _{0}^{\pi} \dfrac{4x\sin x}{1+\cos^{2}x}dx$
$=\displaystyle \int _{0}^{\pi/2} \left[\dfrac{4x\sin x}{1+\cos^{2}x} + \dfrac{4\left(\pi-x\right)\sin x}{1+\cos^{4}x}\right]dx$
$=\displaystyle \int _{0}^{\pi/2}4\pi\left(\dfrac{\sin x}{1+\cos^{2}x}\right) dx =\displaystyle \int _{1}^{0}4\pi\left(\dfrac{1}{1+\cos^{2}x}\right)\left(-d\left(\cos x\right)\right)$
$=\tan^{-1}\left(\cos x\right).4\pi$ for $\cos x=0\rightarrow 1$
$= \pi^{2}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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