int_0^{pi /2} {left( {frac{theta }{{sin theta | vedclass

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(A) Let $I = \int_0^{\pi /2} {\left( {\frac{	heta }{{\sin 	heta }}} ight)^2 d	heta } $. Using integration by parts,let $u = 	heta^2$ and $dv = \

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$\pi \log 2$

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(A) Let $I = \int_0^{\pi /2} {\left( {\frac{	heta }{{\sin 	heta }}} ight)^2 d	heta } $. Using integration by parts,let $u = 	heta^2$ and $dv = \csc^2 	heta \, d	heta$. Then $du = 2	heta \, d	heta$ and $v = -\cot 	heta$. $I = [-	heta^2 \cot 	heta]_0^{\pi/2} + \int_0^{\pi/2} 2	heta \cot 	heta \, d	heta$. Evaluating the boundary term: $\lim_{	heta 	o \pi/2} (-	heta^2 \cot 	heta) = 0$ and $\lim_{	heta 	o 0} (-	heta^2 \cot 	heta) = \lim_{	heta 	o 0} (-	heta^2 \frac{\cos 	heta}{\sin 	heta}) = 0$. So,$I = 2 \int_0^{\pi/2} 	heta \cot 	heta \, d	heta$. Using integration by parts again,let $u = 	heta$ and $dv = \cot 	heta \, d	heta$. Then $du = d	heta$ and $v = \ln(\sin 	heta)$. $I = 2 [	heta \ln(\sin 	heta)]_0^{\pi/2} - 2 \int_0^{\pi/2} \ln(\sin 	heta) \, d	heta$. Since $\ln(\sin(\pi/2)) = \ln(1) = 0$ and $\lim_{	heta 	o 0} 	heta \ln(\sin 	heta) = 0$,the first term is $0$. We know that $\int_0^{\pi/2} \ln(\sin 	heta) \, d	heta = -\frac{\pi}{2} \ln 2$. Therefore,$I = 0 - 2(-\frac{\pi}{2} \ln 2) = \pi \ln 2$.

$\int_0^{\pi /2} {\left( {\frac{\theta }{{\sin \theta }}} \right)^2 d\theta = } $

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