int_0^{pi /2} {xcot x,dx} equals | vedclass

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(B) Let $I = \int_0^{\pi /2} {x\cot x\,dx}$.Using integration by parts,let $u = x$ and $dv = \cot x \, dx$. Then $du = dx$ and $v = \log(\sin x)$.Appl

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$\frac{\pi }{2}\log 2$

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(B) Let $I = \int_0^{\pi /2} {x\cot x\,dx}$.Using integration by parts,let $u = x$ and $dv = \cot x \, dx$. Then $du = dx$ and $v = \log(\sin x)$.Applying the formula $\int u \, dv = uv - \int v \, du$:$I = [x \log(\sin x)]_0^{\pi /2} - \int_0^{\pi /2} \log(\sin x) \, dx$.Evaluating the boundary term:At $x = \pi/2$,$x \log(\sin x) = \frac{\pi}{2} \log(1) = 0$.As $x 	o 0^+$,$x \log(\sin x) 	o 0$ (using $L$'Hopital's rule).Thus,$[x \log(\sin x)]_0^{\pi /2} = 0$.We know that $\int_0^{\pi /2} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2$.Substituting this back into the expression for $I$:$I = 0 - (-\frac{\pi}{2} \log 2) = \frac{\pi}{2} \log 2$.

$\int_0^{\pi /2} {x\cot x\,dx} $ equals

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