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Question Number 174430 by CElcedricjunior last updated on 31/Jul/22

	Answered by som(math1967) last updated on 01/Aug/22
	$I=∫_0 ^π ((xsinx)/(2(1+cos^2 x)))dx =∫_0 ^π (((π−x)sin(π−x))/(2{1+cos^2 (π−x)}))dx =(π/2)∫_0 ^π ((sinxdx)/(1+cos^2 x)) −∫_0 ^π ((xsinx)/(1+cos^2 x))dx ∴I=(π/2)∫_0 ^π ((sinxdx)/(1+cos^2 x)) −I 2I=(π/2)∫_0 ^π ((−d(cosx))/(1+cos^2 x)) 2I=(π/2)[−tan^(−1) cosx]^π _0 2I=(π/2)[−(−(π/4))+((π/4))] I=(π^2 /8)$
	$I = \int_{0}^{π} \frac{x s i n x}{2 (1 + {c o s}^{2} x)} d x$ $= \int_{0}^{π} \frac{(π - x) s i n (π - x)}{2 {1 + {c o s}^{2} (π - x)}} d x$ $= \frac{π}{2} \int_{0}^{π} \frac{s i n x d x}{1 + {c o s}^{2} x} - \int_{0}^{π} \frac{x s i n x}{1 + {c o s}^{2} x} d x$ $∴ I = \frac{π}{2} \int_{0}^{π} \frac{s i n x d x}{1 + {c o s}^{2} x} - I$ $2 I = \frac{π}{2} \int_{0}^{π} \frac{- d (c o s x)}{1 + {c o s}^{2} x}$ $Missing \left or extra \right$ $2 I = \frac{π}{2} [- (- \frac{π}{4}) + (\frac{π}{4})]$ $I = \frac{π^{2}}{8}$
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