<p>Evaluate \(\displaystyle\int_0^1(\arcsin x)^2\,dx\) [JEE Advanced 2007]</p>
Step-by-Step Solution
Key Concept: Let x=sin\theta: \int_0^(\pi/2) \theta^2cos\theta d\theta. IBP twice gives [\theta^2sin\theta-2\thetacos\theta-2sin\theta+2... ]_0^(\pi/2) = \pi^2/4-2.
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<p>Let $x=\sin\theta$, $dx=\cos\theta\,d\theta$. Limits: $\theta:0\to\pi/2$.</p>
<p>$$I=\int_0^{\pi/2}\theta^2\cos\theta\,d\theta$$</p>
<p>IBP: $u=\theta^2$, $dv=\cos\theta\,d\theta$: $=[\theta^2\sin\theta]_0^{\pi/2}-2\int_0^{\pi/2}\theta\sin\theta\,d\theta=\frac{\pi^2}{4}-2\int_0^{\pi/2}\theta\sin\theta\,d\theta$</p>
<p>$\int_0^{\pi/2}\theta\sin\theta\,d\theta=[-\theta\cos\theta]_0^{\pi/2}+\int_0^{\pi/2}\cos\theta\,d\theta=0+1=1$</p>
<p>$$I=\frac{\pi^2}{4}-2\cdot 1=\boxed{\frac{\pi^2}{4}-2}$$</p>
Correct Answer: A