Question

${\int }_0^{2\pi }{\sin }^{4}xdx$

Answer 1

${\int }_0^{2\pi }{\sin }^{4}xdx={\int }_0^{2\pi }({\sin }^{2}x{)}^{2}dx.$

$={\int }_0^{2\pi }{\left(\frac{1-\cos 2x}{2}\right)}^{2}dx$

$={\int }_{2\pi }^0\left[\frac{1-2\cos 2x+{\cos }^{2}2x}{4}\right]dx$

$=\frac{1}{4}{\int }_0^{2\pi }dx-\frac{1}{2}{\int }_0^{2\pi }\cos 2xdx+\frac{1}{4}{\int }_0^{2\pi }{\cos }^{2}2xdx$

$=\frac{1}{4}[x{]}_0^{2\pi }-\frac{1}{2}{\left[\frac{\sin 2x}{2}\right]}_0^{2\pi }+\frac{1}{4}{\int }_0^{2\pi }\left(\frac{1+\cos 4x}{2}\right)dx$

$\frac{1}{4}\left[2x\right]-\frac{1}{4}\left[\sin 4\pi \right]+\frac{1}{8}[x{]}_0^{2\pi }+\frac{1}{8}{\left[\frac{\sin 4x}{4}\right]}_0^{2\pi }$

$=\frac{\pi }{2}-0+\frac{\pi }{4}+(\frac{1}{32}\times 0)+C,c=$ constant of integration

$=\frac{\pi }{2}+\frac{\pi }{4}=\frac{2\pi +\pi }{4}=\frac{3\pi }{4}+C$ Ans