<s>[INST] Find $\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx$ Find $\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx$
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This appears to be an easy problem, but it is consuming a lot of time, I am wondering if an easy way is possible.
WHAT I DID :
Wrote this as $\int \dfrac{1}{4}(\sin^22x)\cos^2x$
And then I wrote $\cos^2x$ in terms of $\cos(2x)$, I get an Integral which is sum of two known integrals, I did them, I got answer $\dfrac{\pi}{16}$
 [/INST] $$ 2\sin x \cos x = \sin 2x$$
$$ 2\cos^2 x -1 = \cos 2x \iff \cos 2x = \frac{ \cos 2x+1}{2}$$
$$\sin^2 x \cos^4 x =\sin^2 x \cos^2 x cos^2 x= 1/8 \sin^2 2x ( \cos 2x+1) = 1/8 (\sin^22x\cos2x+\sin^2 2x)$$
$$\int \sin^2 x \cos^4 x dx = 1/8 (\int \sin^22x\cos2xdx+\int \sin^22x dx)$$
denotation: 
$$ A = \int \sin^22x\cos2xdx $$
$$ B = \int \sin^22x dx$$
For A, Let$ u = \sin2x$ then $du=2\cos2x dx$ (i.e. $\cos xdx = du/2$)
$$A=\int u^2/2 du = u^3/6 = \sin^3 2x/6$$
For B, note that $ 1-2\sin^2\theta = \cos 2 \theta \iff \sin^2 \theta =  (1-cos 2 \theta)/2$ 
$$B=1/2 (\int dx -\int cos 4x dx) =1/2(x-\frac{\sin4x}{4})$$
In a nutshell:
$$\int \sin^2 x \cos^4 x dx = 1/8 (A+B) = 1/48 \sin^3 2x + 1/16 x - 1/64 \sin 4x$$
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