<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] With even powers of the trig functions use the half-angle identities until you have a an odd power.
$\sin^2 x = \frac 12 (1-\cos 2x)\\
\cos^2 x = \frac 12 (1+\cos 2x)$
$(\sin^2 x)(\cos^4 x) = \frac {1}{8}(1-\cos 2x)(1+cos 2x)(1+\cos 2x)\\
\frac18(1-\cos^2 2x)(1+\cos 2x)\\
\frac 18 (1 +\cos 2x - \cos^2 2x - \cos^3 2x)\\
\frac 18 (1 +\cos 2x - \frac 12 (1+\cos 4x)  - \cos 2x(1-\sin^2 2x))\\
\frac 1{16} (1  - \cos 4x  - 2\sin^2 2x\cos 2x)$
And each of those terms is straight forward to integrate.
Alternatively, you can use some complex analysis and say:
$\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\
\cos x = \frac {e^{ix} + e^{-ix}}{2}$
$(\sin^2 x)(\cos^4 x) = \frac {(e^{2ix} -2 + e^{-2ix})(e^{4ix} + 4e^{2ix} + 6 + 4e^{-2ix} + e^{-4ix})}{-64}\\
\frac {e^{6ix} + 2e^{4ix} - e^{2ix} -4 - e^{-2ix} +2 e^{-4ix} + e^{-6ix}}{-64}\\
\frac {-\cos 6x - 2\cos 4x + \cos 2x  + 2}{32}$
And again not to bad to integrate.
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