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 $\dfrac{\pi}{16}$

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
