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Find $\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx$

$ $

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 $\dfrac{\pi}{16}$

Fusion2
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    $\sin^2(x) = \frac{1-\cos(2x)}{2}$, $\cos^4(x) = \left ( \frac{1+\cos(2x)}{2} \right )^2$. After you expand and combine like terms you should have $a + b \cos(2x) + c \cos^2(2x) + d \cos^3(2x)$. The first two are easy enough. For the third, you can reduce the power again using the same formula. The fourth you can handle by using the Pythagorean identity to change two of the cosine factors into sine factors and then use a simple substitution. Or you can write $\sin^2(x) \cos^4(x) = \left ( \frac{e^{ix} - e^{-ix}}{2i} \right )^2 \left ( \frac{e^{ix} + e^{-ix}}{2} \right )^4$. – Ian Sep 14 '14 at 02:23
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    Write $\sin^2x=1-\cos^2x$, and then break it up into a sum of two Wallis integrals. – Lucian Sep 14 '14 at 03:34
  • $$ \small{\int^\pi_0\sin^2{x}\cos^4{x}\ {\rm d}x =\int^{\frac{\pi}{2}}0\sin^2{x}\cos^4{x}\ {\rm d}x+\int^\pi{\frac{\pi}{2}}\sin^2{x}\cos^4{x}\ {\rm d}x =\frac{1}{2}B\left(\frac{3}{2},\frac{5}{2}\right)-\int^{-\frac{\pi}{2}}_{0}\sin^4{x}\cos^2{x}\ {\rm d}x =B\left(\frac{3}{2},\frac{5}{2}\right) =\frac{1}{3!}\frac{\sqrt{\pi}}{2}\frac{3\sqrt{\pi}}{4} =\frac{\pi}{16}} $$ – SuperAbound Sep 14 '14 at 06:43
  • See this answer for an explanation of a method for finding the indefinite integrals. I encourage you to add reduction to Wallis integrals (follow Lucian's link) to your repertoire. – Jyrki Lahtonen Jun 30 '16 at 11:36

4 Answers4

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$$\int_{0}^{\frac{\pi}{2}}\sin^{2\alpha-1}x\,\cos^{2\beta-1}x\,dx=\frac{\Gamma(\alpha)\Gamma(\beta)}{2\Gamma(\alpha+\beta)}$$ $$\int_0^{\pi}\sin^2x\cos^4x\hspace{1mm}dx=2\int_0^{\frac{\pi}{2}}\sin^2x\cos^4x\hspace{1mm}dx=2\times\frac{\Gamma(\frac32)\Gamma(\frac52)}{2\Gamma(4)}=\frac{3\pi}{48}=\frac{\pi}{16}$$

Behrouz Maleki
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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".

K7PEH
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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

Claude Leibovici
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$$ 2\sin x \cos x = \sin 2x$$ $$ 2\cos^2 x -1 = \cos 2x \iff \cos 2x = \frac{ \cos 2x+1}{2}$$ $$\sin^2 x \cos^4 x =\sin^2 x \cos^2 x cos^2 x= 1/8 \sin^2 2x ( \cos 2x+1) = 1/8 (\sin^22x\cos2x+\sin^2 2x)$$ $$\int \sin^2 x \cos^4 x dx = 1/8 (\int \sin^22x\cos2xdx+\int \sin^22x dx)$$

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: $$\int \sin^2 x \cos^4 x dx = 1/8 (A+B) = 1/48 \sin^3 2x + 1/16 x - 1/64 \sin 4x$$

Zau
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