AP EAMCET · Maths · Definite Integration
\(\int_0^\pi \frac{\theta \sin \theta}{1+\cos ^2 \theta} d \theta\) is equal to
- A \(\frac{\pi^2}{2}\)
- B \(\frac{\pi^2}{3}\)
- C \(\pi^2\)
- D \(\frac{\pi^2}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi^2}{4}\)
Step-by-step Solution
Detailed explanation
\(=\int_0^\pi \frac{(\pi-\theta) \sin (\pi-\theta)}{1+\cos ^2(\pi-\theta)} d \theta\) On adding Eqs. (i) and (ii), we get \(2 I=\int_0^\pi \frac{\pi \sin \theta}{1+\cos ^2 \theta} d \theta\) Let \(\cos \theta=t \Rightarrow-\sin \theta d \theta=d t\)…
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