<s>[INST] 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.
 [/INST] You can calculate it directly without substitution as follows:
$$\sin^2x\cos^4x=(\sin x\cos x)^2\cos^2x =\frac 14\sin^22x\cdot \frac 12(1+\cos 2x)$$
$$= \frac 18 \cdot\frac 12(1-\cos 4x) +\frac 18\sin^2 2x\cos 2x$$
Hence,
$$\int \sin^2x\cos^4x\; dx = \frac x{16}-\frac{\sin 4x}{64} + \frac{\sin^3 2x}{48}+c$$
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