Q54 / 170 · Manipal (MET) 2017 · Mathematics

$\int_0^1 \frac{d}{d x}\left[\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right] d x$ is equal to

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

Solution

Let $I=\int_0^1 \frac{d}{d x}\left[\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right] d x$ $\begin{aligned} \therefore I & =\int_0^1 \frac{d}{d x}\left[\sin ^{-1}(\sin 2 \theta)\right] d x \quad(\text { put } x=\tan \theta) \\ & =2 \int_0^1 \frac{d}{d x}\left[\tan ^{-1} x\right] d x \\ & =2\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \\ & =2 \cdot \frac{\pi}{4}=\frac{\pi}{2} \end{aligned}$

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

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