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.
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.
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?
"Hint":
$\int \sin^4(x)dx=\int [\sin^2(x)]^2dx=\frac{1}{4}\int(1-\cos(2x))^2dx$
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 $\,\displaystyle{\frac{3\pi}{8}}\,$