Q87 / 150 · MHT CET 2022 (05 Aug Shift 1) · Mathematics

$\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} d x$ has the value

  • A
    $-\frac{\pi}{4}$
  • Correct answer
    B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{2}$
  • D
    $0$

Solution

$\begin{aligned} & \int_0^{\pi / 2} \frac{\sin x}{1+\cos ^2 \mathrm{x}} \mathrm{dx}=-\int_1^0 \frac{\mathrm{dt}}{1+\mathrm{t}^2}[\text { let } \cos \mathrm{x}=\mathrm{t}] \\ & =-\left[\tan ^{-1} \mathrm{t}\right]_1^0 \\ & =-\left(0-\frac{\pi}{4}\right) \\ & =\frac{\pi}{4}\end{aligned}$

Hence, the correct answer is option: 'B'.

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