If int frac{cos ^3 x}{sin ^2 x+sin ^4 x} d x= | vedclass

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(C) Let $I = \int \frac{\cos ^3 x}{\sin ^2 x+\sin ^4 x} d x$. Substitute $u = \sin x$,then $du = \cos x dx$. The integral becomes $I = \int \frac{\cos

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$\frac{\pi}{2}$

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(C) Let $I = \int \frac{\cos ^3 x}{\sin ^2 x+\sin ^4 x} d x$. Substitute $u = \sin x$,then $du = \cos x dx$. The integral becomes $I = \int \frac{\cos ^2 x}{\sin ^2 x+\sin ^4 x} \cos x dx = \int \frac{1-u^2}{u^2+u^4} du$. $I = \int \frac{1-u^2}{u^2(1+u^2)} du$. Using partial fractions,$\frac{1-u^2}{u^2(1+u^2)} = \frac{A}{u} + \frac{B}{u^2} + \frac{Cu+D}{1+u^2}$. Solving for constants,we get $\frac{1-u^2}{u^2(1+u^2)} = \frac{1}{u^2} - \frac{2}{1+u^2}$. Thus,$I = \int (u^{-2} - 2(1+u^2)^{-1}) du = -u^{-1} - 2 	an^{-1}(u) + c$. Substituting back $u = \sin x$,$I = -\frac{1}{\sin x} - 2 	an^{-1}(\sin x) + c = c - \operatorname{cosec} x - 2 	an^{-1}(\sin x)$. Comparing with the given form $c - \operatorname{cosec} x - f(x)$,we have $f(x) = 2 	an^{-1}(\sin x)$. Therefore,$f\left(\frac{\pi}{2}ight) = 2 	an^{-1}(\sin \frac{\pi}{2}) = 2 	an^{-1}(1) = 2 	imes \frac{\pi}{4} = \frac{\pi}{2}$.

If $\int \frac{\cos ^3 x}{\sin ^2 x+\sin ^4 x} d x=c-\operatorname{cosec} x-f(x)$,then $f\left(\frac{\pi}{2}\right)=$

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