Evaluate _ 0 ^ 2 x 1+ ^ 2 x d x .
Mathematics · JEE Main · NTA Exams — Integral Calculus
Evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x\].
- \(\frac{\pi}{4}\)
- \(\frac{\pi}{2}\)
- \(\frac{\pi}{8}\)
- \(\frac{\pi}{6}\)
Answer
(A) 4
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