<p>Evaluate \(\displaystyle\int_0^{\pi/2}x\cos x\,dx\) [JEE Main 2018]</p>
Step-by-Step Solution
Key Concept: IBP: u=x, dv=cosx dx. [x sinx]_0^(\pi/2) - \int_0^(\pi/2) sinx dx = \pi/2 - [-cosx]_0^(\pi/2) = \pi/2 - 1.
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<p>IBP ($u=x, dv=\cos x\,dx$): $[x\sin x]_0^{\pi/2}-\int_0^{\pi/2}\sin x\,dx=\frac{\pi}{2}-[-\cos x]_0^{\pi/2}=\frac{\pi}{2}-(0+1)=\frac{\pi}{2}-1$</p>
<p>Wait -- answer B = 1. Recheck: $=\frac{\pi}{2}-(-\cos(\pi/2)+\cos 0)=\frac{\pi}{2}-(0+1)=\frac{\pi}{2}-1\approx 0.571$. So answer is A=\pi/2-1. Accept A.</p>
Correct Answer: B