Question

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

• A. $\frac{\pi^{2}}{2}$
• B. $\frac{\pi^{2}}{8}$
• C. $\frac{\pi^{2}}{4}$
• D. None of these

Answer 1

Answer: (C)

$\frac{\pi^{2}}{4}$

Answer 2

Let I = $\int_{0}^{\pi}\frac{x\sin x}{1 + \cos^{2}x}$dx … (1)

= $\int_{0}^{\pi}\frac{(\pi - x)\sin(\pi - x)}{1 + \cos^{2}(\pi - x)}$ (By Prop. IV)

= $\int_{0}^{\pi}\frac{(\pi - x)\sin x}{1 + \cos^{2}x}$ … (2)

Adding (1) and (2), we get

2I = p $\int_{0}^{\pi}\frac{\sin x}{1 + \cos^{2}x}$dx

Put cos x = t

When x = 0 Ž t = 1

x = p Ž t = –1

\ 2I = –p$\int_{1}^{–1}\frac{dt}{1 + t^{2}}$

= p$\int_{–1}^{1}\frac{dt}{1 + t^{2}}$

= p tan^–1$\left. \ t \right|_{- 1}^{1}$

= p {tan^–1 1 – tan^–1 (–1)}

= p {2 tan^–1 1}

\ I = p tan^–1 1 = p . $\frac{\pi}{4}$

Hence I = $\frac{\pi^{2}}{4}$.