Evaluate: $\int_{0}^{π} \frac{xsinx}{1 + \cos^{2} x} dx$

Solution

$I = \int_{0}^{π} \frac{xsinx}{1 + \cos^{2} x} dx . . . . . . . . . . . . . (1) I = \int_{0}^{π} \frac{(π - x) \sin (π - x)}{1 + \cos^{2} (π - x)} dx I = \int_{0}^{π} \frac{(π - x) \sin x}{1 + \cos^{2} x} dx I = \int_{0}^{π} \frac{π \sin x}{1 + \cos^{2} x} dx - \int_{0}^{π} \frac{xsinx}{1 + \cos^{2} x} dx . . . . . . . . . . (2)$

Adding (1) and (2), we get:

$2 I = \int_{0}^{π} \frac{π sinx}{1 + \cos^{2} x} dx Now, let cosx = t \Rightarrow - sinx dx = dt When x = π, t = cosπ = - 1 When x = 0, t = \cos 0 = 1 2 I = \int_{1}^{- 1} \frac{π - dt}{1 + t^{2}} 2 I = π \int_{1}^{- 1} (\frac{1}{1 + t^{2}}) dt 2 I = π {[\tan^{- 1} t]}_{1}^{- 1} 2 I = π [\tan^{- 1} 1 - \tan^{- 1} - 1] 2 I = π (\frac{π}{4} - (- \frac{π}{4})) 2 I = \frac{π^{2}}{2} ∴ I = \frac{π^{2}}{4}$

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Evaluate: $\int_{0}^{π} \frac{xsinx}{1 + \cos^{2} x} dx$