Integral of $\int \sin^2x\cos^4xdx$ $$\int \sin^2x\cos^4xdx$$
I tried 
$$I = \int (1-\cos^2x)\cos^4xdx = \int \frac{\sec^2x-1}{\sec^6x}dx = \int \frac{\tan^2x}{\sec^6x}dx$$
Take $\tan x = t \implies \sec^2xdx = dt$
$$I = \int \frac{t^2}{(t^2+1)^4}dt$$
And I could not proceed further from here.

$$I=\int \sin^2 x \cos^4 x dx =\frac{1}{8} \int \sin^2 2x (1+\cos 2x) dx=\frac{1}{8}\int \sin ^2 2x dx+\frac{1}{8}\int (t^2/2)  dt$$
Here $\sin 2x=t$
$$\implies I=\frac{1}{16}\int (1-\cos 4x) dx +\frac{(\sin 2x)^3}{48} =\frac{1}{16} x-\frac{1}{64} \sin 4x+\frac{(\sin 2x)^3}{48} $$
