int frac{cos ^3 x+cos ^5 x}{sin ^2 x+sin ^4 x | vedclass

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(C) Let $I = \int \frac{\cos ^3 x+\cos ^5 x}{\sin ^2 x+\sin ^4 x} d x$. Substitute $\sin x = t$,then $\cos x d x = d t$. $I = \int \frac{\cos ^2 x(1+\

Vedclass · Accepted Answer

$\sin x-2(\sin x)^{-1}-6 	an ^{-1}(\sin x)+c$

Vedclass · Answer

(C) Let $I = \int \frac{\cos ^3 x+\cos ^5 x}{\sin ^2 x+\sin ^4 x} d x$. Substitute $\sin x = t$,then $\cos x d x = d t$. $I = \int \frac{\cos ^2 x(1+\cos ^2 x)}{\sin ^2 x(1+\sin ^2 x)} \cos x d x = \int \frac{(1-t^2)(1+1-t^2)}{t^2(1+t^2)} d t = \int \frac{(1-t^2)(2-t^2)}{t^2(1+t^2)} d t$. $I = \int \frac{t^4-3t^2+2}{t^2(1+t^2)} d t$. Using partial fractions or algebraic manipulation: $\frac{t^4-3t^2+2}{t^2(1+t^2)} = \frac{t^2(t^2+1)-4t^2+2}{t^2(1+t^2)} = 1 + \frac{-4t^2+2}{t^2(1+t^2)}$. $\frac{-4t^2+2}{t^2(1+t^2)} = \frac{A}{t^2} + \frac{B}{1+t^2} \Rightarrow -4t^2+2 = A(1+t^2) + Bt^2$. For $t^2=0$,$A=2$. For $t^2=-1$,$-B = 6 \Rightarrow B=-6$. So,$I = \int (1 + \frac{2}{t^2} - \frac{6}{1+t^2}) d t = t - \frac{2}{t} - 6 	an^{-1} t + c$. Substituting $t = \sin x$,we get $I = \sin x - 2(\sin x)^{-1} - 6 	an^{-1}(\sin x) + c$.

$\int \frac{\cos ^3 x+\cos ^5 x}{\sin ^2 x+\sin ^4 x} d x=$

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