$\int_0^\pi \frac{\theta \sin \theta}{1+\cos ^2 \theta} d \theta$ is equal to

• A

  $\frac{\pi^2}{2}$
• B

  $\frac{\pi^2}{3}$
• C

  $\pi^2$
• Correct answer
  D

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

Solution

$=\int_0^\pi \frac{(\pi-\theta) \sin (\pi-\theta)}{1+\cos ^2(\pi-\theta)} d \theta$
On adding Eqs. (i) and (ii), we get $2 I=\int_0^\pi \frac{\pi \sin \theta}{1+\cos ^2 \theta} d \theta$ Let $\cos \theta=t \Rightarrow-\sin \theta d \theta=d t$ $\begin{aligned} \therefore \quad 2 I & =-\pi \int_1^{-1} \frac{1}{1+t^2} d t \\ & =2 \pi \int_0^1 \frac{1}{1+t^2} d t \\ & =2 \pi\left[\tan ^{-1} t\right]_0^1=2 \pi \cdot \frac{\pi}{4} \end{aligned}$ $\Rightarrow \quad I=\frac{\pi^2}{4}$

Hence, the correct answer is option: 'D'.