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GATE 2017 CE (CE2) – Question 37

Engineering Mathematics · Calculus: Integrals, partial and total derivatives, gradient, divergence, curl, line, surface and volume integrals · 2 marks · Multiple choice

Evaluate $I=\int_0^1\frac{(\sin^{-1}x)^2}{\sqrt{1-x^2}}\,dx$.

  1. $\pi^3/24$
  2. $\pi^3/12$
  3. $\pi^3/48$
  4. $\pi^3/64$

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Correct answer: (A) $\pi^3/24$

Explanation

Put $u=\sin^{-1}x$, so $du=dx/\sqrt{1-x^2}$. The limits become 0 and π/2. Therefore $$I=\int_0^{\pi/2}u^2\,du=\left[\frac{u^3}{3}\right]_0^{\pi/2}=\frac{\pi^3}{24}.$$ Answer A.