Q110 / 170 · Manipal (MET) 2021 · Mathematics

$\int_0^1 \sin ^{-1} x d x$ is

  • Correct answer
    A
    $\frac{\pi}{2}-1$
  • B
    1
  • C
    $\pi$
  • D
    0

Solution

$\begin{aligned} & \text { Let } I=\int_0^1 \sin ^{-1} x d x \\ & \qquad=\left[x \sin ^{-1} x\right]_0^1-\int_0^1 \frac{x}{\sqrt{1-x^2}} d x\end{aligned}$ Put $1-x^2=t^2 \Rightarrow-2 x d x=2 t d t$ $\therefore \quad I=\left[\frac{\pi}{2}-0\right]+\int_1^0 \frac{t d t}{t}$ $=\frac{\pi}{2}+\int_1^0 d t=\frac{\pi}{2}+[t]_1^0=\frac{\pi}{2}-1$

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

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