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Question: Answered & Verified by Expert
$\int_0^\pi \frac{\theta \sin \theta}{1+\cos ^2 \theta} d \theta$ is equal to
MathematicsDefinite IntegrationAP EAMCETAP EAMCET 2005
Options:
  • A $\frac{\pi^2}{2}$
  • B $\frac{\pi^2}{3}$
  • C $\pi^2$
  • D $\frac{\pi^2}{4}$
Solution:
1302 Upvotes Verified Answer
The correct answer is: $\frac{\pi^2}{4}$


$=\int_0^\pi \frac{(\pi-\theta) \sin (\pi-\theta)}{1+\cos ^2(\pi-\theta)} d \theta$

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

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