<p>Evaluate \(\displaystyle\int_0^{\pi/2}x\cot x\,dx\). [JEE Advanced 2011]</p>
Step-by-Step Solution
Key Concept: IBP: u=x, dv=cotx dx. [x ln sinx]_0^(\pi/2) - \int_0^(\pi/2) ln(sinx)dx = 0 - (-\pi/2 \cdot ln2) = \pi/2 \cdot ln2.
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<p>IBP: $u=x, dv=\cot x\,dx\Rightarrow v=\ln\sin x$.</p>
<p>$\int_0^{\pi/2}x\cot x\,dx=[x\ln\sin x]_0^{\pi/2}-\int_0^{\pi/2}\ln\sin x\,dx$</p>
<p>At $x=\pi/2$: $(\pi/2)\ln 1=0$. At $x=0$: $x\ln\sin x\to x\ln x\to 0$.</p>
<p>$$=0-\left(-\frac{\pi}{2}\ln 2\right)=\frac{\pi}{2}\ln 2=\boxed{\frac{\pi\ln 2}{2}}$$</p>
Correct Answer: A