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How can I solve this integral

$$\int_0^\pi \sin^4{x}\,dx\;\text{?}$$

Is beta function used to solve it? I want the way.

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You don't solve integrals, you evaluate them. – Stefan Smith Oct 16 '12 at 21:26
up vote 2 down vote accepted

Put $\,\displaystyle{I:=\int\sin^4x\,dx}\,$

Now, we have that$$\sin^4x=\sin^2x(1-\cos^2x):$$

By parts the second integral:

$$u=\cos x\;,\;u'=-\sin x$$

$$v'=\sin^2x\cos x\;,\;v=\frac{1}{3}\sin^3x\Longrightarrow$$

$$\int\sin^2x\cos^2x\,dx=\frac{1}{3}\sin^3x\cos x+\frac{1}{3}I$$

So we get that:

$$I=\int\sin^2x\,dx-\frac{1}{3}\sin^3x\cos x-\frac{1}{3}I\Longrightarrow$$

$$\Longrightarrow \frac{4}{3}I=\frac{x-\sin x\cos x}{2}+\frac{1}{3}\sin^3x\cos x\Longrightarrow$$

$$\Longrightarrow I=\frac{3}{8}(x-\sin x\cos x)+\frac{1}{4}\sin^3x\cos x+K$$

Now do the above with your definite integral: the result is $\,\displaystyle{\frac{3\pi}{8}}\,$

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$$\int \sin^4(x)dx=\int [\sin^2(x)]^2dx=\frac{1}{4}\int(1-\cos(2x))^2dx$$

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To continue, you FOIL out the square and then use $\cos^2(2x)=\frac{1}{2}(1+2\cos(4x))$... – Stefan Smith Oct 16 '12 at 21:48

Beta function formula (see ) $$ \mathrm{B}(x,y) = 2 \int_0^{\pi/2} (\sin\theta)^{2x-1}(\cos\theta)^{2y-1}d\theta $$ Perhaps that is what you need to use?

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