Mathematics

# $\displaystyle\int^{\pi}_{-\pi}\dfrac{2x(1+\sin x)}{1+\cos^2x}dx$.

##### SOLUTION
$\displaystyle I = \int_{-\pi}^{\pi} \cfrac{2x}{1+\cos^2x}dx + \int_{-\pi}^{\pi} \cfrac{2x\sin x}{1+\cos^2x} dx$
$I = \displaystyle \int_0^{\pi} 2\cfrac{(\pi-x)\sin x}{1 + \cos^2x} dx$
$2I = \displaystyle 2\pi\int_0^{\pi}\cfrac{\sin x}{1+\cos^2x}$
$I = \displaystyle \pi \int_{-1}^1 \cfrac{dt}{1+t^2}$
$I = \displaystyle \pi \Bigg[\tan^{-1} t\Bigg]_{-1}^1 = \cfrac{\pi^2}{2}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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