MCQ
$\int_0^{\pi / 2} \frac{\sin x}{1+\cos ^2 x} d x$ has the value
- A
- B$\pi / 2$
- C$\pi / 4$
- D$-\pi / 4$
(c) : Let $I=\int_0^{\pi / 2} \frac{\sin x}{1+\cos ^2 x} d x$
Let $\cos x=t \Rightarrow-\sin x d x=d t$, when $x=0, t=1, x=\pi / 2$, $t=0$
$\Rightarrow I=-\int_1^0 \frac{d t}{1+t^2}=\left[-\tan ^{-1} t\right]_1^0=\tan ^{-1} 1=\frac{\pi}{4}$
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