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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.

  • 1
    You don'$t$ solve integrals, you evaluate them.2012-10-16

3 Answers 3

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Beta function formula (see http://en.wikipedia.org/wiki/Beta_function ) $ \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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"Hint":

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

  • 0
    To continue, you FOIL out the square and then use $\cos^2(2x)=\frac{1}{2}(1+2\cos(4x))$...2012-10-16
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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:

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

$\Longrightarrow \frac43I=\frac{x-\sin x\cos x}2-\frac{1}{3}\sin^3x\cos x\Longrightarrow$

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

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