MCQ
$\int_0^{\pi /2} {\frac{{\sin x}}{{1 + {{\cos }^2}x}}\,dx} =$
- A$\pi /2$
- ✓$\pi /4$
- C$\pi /3$
- D$\pi /6$
Put $\cos x = t$ ==> $ - \sin x\,dx = dt$
Then $I = \int_1^0 {\frac{{ - dt}}{{1 + {t^2}}} = \int_0^1 {\frac{{dt}}{{1 + {t^2}}} = [{{\tan }^{ - 1}}t]_0^1 = \frac{\pi }{4}} } $.
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