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 $\dfrac{\pi}{16}$
 A: $$\int_{0}^{\frac{\pi}{2}}\sin^{2\alpha-1}x\,\cos^{2\beta-1}x\,dx=\frac{\Gamma(\alpha)\Gamma(\beta)}{2\Gamma(\alpha+\beta)}$$
$$\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx=2\int_0^{\frac{\pi}{2}}\sin^2x\cos^4x\hspace{1mm}dx=2\times\frac{\Gamma(\frac32)\Gamma(\frac52)}{2\Gamma(4)}=\frac{3\pi}{48}=\frac{\pi}{16}$$
A: You got the right answer.  Not all integrals are easy to solve.  Some are tricky and take time.  If you want to see some examples of this, get the new book by Paul J. Nahin "Inside Interesting Integrals".
A: First look at the antiderivative $$I=\int\sin^2(x)\cos^4(x)\hspace{1mm}dx$$ and now use $$\sin^2(x)=\frac{1-\cos(2x)}{2}$$ $$\cos^2(x)=\frac{\cos(2x)+1}{2}$$ so $$\cos^4(x)=\Big(\frac{\cos(2x)+1}{2}\Big)^2=\frac{1}{4}\Big(\cos^2(2x)+2 \cos(2x)+1\Big)$$ $$\cos^4(x)=\frac{1}{4}\Big(\frac{\cos(4x)+1}{2}+2 \cos(2x)+1\Big)$$ $$\cos^4(x)=\frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x)+\frac{3}{8})$$ So, now $$sin^2(x)\cos^4(x)=\frac{1-\cos(2x)}{2}\Big(\frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x)+\frac{3}{8}\Big)$$ Develop and as a result you will have simple cosines and products of cosines; now remember and use the fact that $$\cos(a)\cos(b)=\frac{1}{2}\Big(\cos(a+b)+\cos(a-b)\Big)$$ Normally, you will end with $$\sin^2(x)\cos^4(x)=\frac{1}{32} \cos (2 x)-\frac{1}{16} \cos (4 x)-\frac{1}{32} \cos (6 x)+\frac{1}{16}$$ and so $$I=\frac{x}{16}+\frac{1}{64} \sin (2 x)-\frac{1}{64} \sin (4 x)-\frac{1}{192} \sin (6 x)$$ For the given integration bounds only the first term has to be used and the result is $\frac{\pi}{16}$.
In my opinion, it is easier to use Euler formula $$\cos x=\frac{e^{ix}+e^{-ix}}2,\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$ so $$\sin^2(x)=-\frac{1}{4} e^{-2 i x}-\frac{1}{4} e^{2 i x}+\frac{1}{2}$$ $$\cos^4(x)=\frac{1}{4} e^{-2 i x}+\frac{1}{4} e^{2 i x}+\frac{1}{16} e^{-4 i x}+\frac{1}{16}
   e^{4 i x}+\frac{3}{8}$$ $$\sin^2(x)\cos^4(x)=\frac{1}{64} e^{-2 i x}+\frac{1}{64} e^{2 i x}-\frac{1}{32} e^{-4 i x}-\frac{1}{32}
   e^{4 i x}-\frac{1}{64} e^{-6 i x}-\frac{1}{64} e^{6 i x}+\frac{1}{16}$$ $$\sin^2(x)\cos^4(x)=\frac{\cos(2x)}{32}-\frac{\cos(4x)}{16}-\frac{\cos(6x)}{32}+\frac{1}{16}$$ and the integration become extremely simple
A: $$ 2\sin x \cos x = \sin 2x$$
$$ 2\cos^2 x -1 = \cos 2x \iff \cos 2x = \frac{ \cos 2x+1}{2}$$
$$\sin^2 x \cos^4 x =\sin^2 x \cos^2 x cos^2 x= 1/8 \sin^2 2x ( \cos 2x+1) = 1/8 (\sin^22x\cos2x+\sin^2 2x)$$
$$\int \sin^2 x \cos^4 x dx = 1/8 (\int \sin^22x\cos2xdx+\int \sin^22x dx)$$
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:
$$\int \sin^2 x \cos^4 x dx = 1/8 (A+B) = 1/48 \sin^3 2x + 1/16 x - 1/64 \sin 4x$$
