$\int {\sin }^{4}xdx=$

$\int {\sin }^{4}x dx$    $\left(\because {\sin }^{2}x=\frac{1-\cos 2x}{2}\right)$

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

$=\frac{1}{4}\int \left[1-2\cos \left(2x\right)+{\cos }^2\left(2x\right)\right]dx$     $\left[\because {\cos }^2\theta =\frac{{\displaystyle 1+\cos \left(2\theta \right)}}{{\displaystyle 2}}\right]$

$=\frac{1}{4}\left[x-\frac{{\displaystyle 2\sin \left(2x\right)}}{{\displaystyle 2}}+\int \frac{{\displaystyle 1+\cos \left(4x\right)}}{{\displaystyle 2}}dx\right]$

$=\frac{1}{4}\left[x-\sin \left(2x\right)+\frac{{\displaystyle 1}}{{\displaystyle 2}}\left(x+\frac{{\displaystyle \sin \left(4x\right)}}{{\displaystyle 4}}\right)\right]+c$

$=\frac{1}{4}x-\frac{1}{4}\sin \left(2x\right)+\frac{1}{8}x+\frac{1}{32}\sin \left(4x\right)+c$

$=\frac{3}{8}x+\frac{1}{32}\sin 4x-\frac{1}{4}\sin 2x+c$