∫ cos 3 x + cos 5 x sin 2 x + sin 4 x d x =

Let,

$I=\int \left(\frac{{\cos }^{3}x+{\cos }^{5}x}{{\sin }^{2}x+{\sin }^{4}x}\right)dx$

    $=\int \left\{\frac{{\cos }^{3}x\left(1+{\cos }^{2}x\right)}{{\sin }^{2}x\left(1+{\sin }^{2}x\right)}\right\}dx$

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

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

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

Let $\sin x=t\Rightarrow \cos xdx=dt$ 

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

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

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

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

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

          $=t-\frac{2}{t}-6{\tan }^{-1}t+c$

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