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

$1={\displaystyle \int \frac{\left({\cos }^{2}x+{\cos }^{4}x\right)\cos xdx}{{\sin }^{2}x+{\sin }^{4}x}}$

$={\displaystyle \int \frac{1-t^2+{\left(1-t^2\right)}^2}{t^2+t^4}}dt,$where $\sin x=t$

$={\displaystyle \int \frac{t^4-3t^2+2}{t^2\left(1+t^2\right)}}dt$

$=2{\displaystyle \int \frac{\left(t^4+t^2\right)-4\left(t^2+1\right)+6}{t^2\left(t^2+1\right)}}dt$

$={\displaystyle \int \left(1-\frac{4}{t^2}+6\left(\frac{1}{t^2}-\frac{1}{1+t^2}\right)\right)dt}$

$=t+\frac{-2}{t}-6{\tan }^{-1}\left(t\right)+c$

$=\sin x-2{\left(\sin x\right)}^{-1}-6{\tan }^{-1}\left(\sin x\right)+c$