The value of $\int\frac{\cos^3x+\cos^5x}{\sin^2x+\sin^4x}dx$ is:

$\sin x-2(\sin x)^{-1}-6\tan^{-1}(\sin x)+C$

Putting sin x = t in the given integral, we get: $\int\frac{1-t^2+(1-t^2)^2}{t^2+t^4}dt=\int\frac{(1-t^2)(2-t^2)}{t^2+t^4}dt$

$=\int(1+\frac{2}{t^2}-\frac{6}{1+t^2})dt=t-\frac{2}{t}-6\tan^{-1}(t)+C=\sin x-2(\sin x)^{-1}-6\tan^{-1}(\sin x)+C$