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integrate x sinx / (1 + cos²x) dx

$I = \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} ⅆ x .................. (i) = \int_{0}^{π} \frac{(π - x) \sin (π - x)}{1 + \cos^{2} (π - x)} ⅆ x [U \sin g \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{π} \frac{(π - x) \sin x}{1 + \cos^{2} x} ⅆ x .................. (i i) A d d i n g (i) & (i i), w e g e t, 2 I = \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} ⅆ x + \int_{0}^{π} \frac{(π - x) \sin x}{1 + \cos^{2} x} ⅆ x = \int_{0}^{π} \frac{(x + π - x) \sin x}{1 + \cos^{2} x} ⅆ x 2 I = \int_{0}^{π} \frac{π \sin x}{1 + \cos^{2} x} ⅆ x I = \frac{π}{2} \int_{0}^{π} \frac{\sin x}{1 + \cos^{2} x} ⅆ x l e t \cos x = t, - \sin x d x = d t A l s o, x = 0, t = \cos 0 = 1 a n d x = π, t = \cos π = - 1 I = \frac{π}{2} \int_{1}^{- 1} \frac{- 1}{1 + t^{2}} ⅆ t I = - \frac{π}{2} {[\tan^{- 1} t]}_{1}^{- 1} = - \frac{π}{2} [\tan^{- 1} (- 1) - \tan^{- 1} (1)] = - \frac{π}{2} [- \frac{π}{4} - \frac{π}{4}] = \frac{π}{2} \times \frac{2 π}{4} I = \frac{π^{2}}{4}$

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