0-pi-x-dx-1-sin-2-x-

Question Number 120688 by bobhans last updated on 01/Nov/20

\underset{0}{\int^{π}} \frac{x dx}{1 + \sin^{2} x}

Answered by john santu last updated on 02/Nov/20

ψ = \underset{0}{\int^{π}} \frac{x d x}{1 + \sin^{2} x}; [l e t x = π - w]

ψ = \underset{π}{\int^{0}} \frac{(π - x) (- d w)}{1 + \sin^{2} {(π - w)}^{2}} = \underset{0}{\int^{π}} \frac{π - w}{1 + \sin^{2} w} d w

a d d i n g t h e e q u a t i o n g i v e

2 ψ = \underset{0}{\int^{π}} \frac{π}{1 + \sin^{2} x} d x

ψ = \frac{1}{2} \underset{0}{\int^{π}} \frac{π}{1 + \sin^{2} x} d x = \frac{π}{2} \underset{0}{\int^{π}} \frac{cosec^{2} x d x}{cosec^{2} x + 1}

ψ = \frac{π}{2} \underset{0}{\int^{π}} \frac{cosec^{2} x d x}{(1 + \cot^{2} x) + 1} = \frac{π}{2} \underset{- \infty}{\int^{\infty}} \frac{1 d t}{2 + t^{2}}; w h e r e t = \cot x

ψ = \frac{π}{2} . \frac{1}{\sqrt{2}} arc \tan (\frac{t}{\sqrt{2}}) ∣_{- \infty}^{\infty}

ψ = \frac{π}{2 \sqrt{2}} [\frac{π}{2} - (- \frac{π}{2})] = \frac{π^{2}}{2 \sqrt{2}} . ▴