Question

Find the value of ${\int }_0^{2\pi }{\sin }^{2}x\cdot {\cos }^{4}xdx$.

Answer 1

Let $I={\int }_0^{2\pi }{\sin }^{2}x{\cos }^{4}xdx=2{\int }_0^{\pi }{\sin }^{2}x{\cos }^{4}xdx$, since $f(2\pi -x)=f\left(x\right)$

$\Rightarrow I=2\left(2\right){\int }_0^{\frac{\pi }{2}}{\sin }^{2}x{\cos }^{4}xdx=4{\int }_0^{\frac{\pi }{2}}{\sin }^{2}x{\cos }^{4}xdx$, since $f(\pi -x)=f\left(x\right)$

$\Rightarrow I=4{\int }_0^{\frac{\pi }{2}}{\sin }^{2}x{\cos }^{4}xdx$.......(1)

Now using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$\Rightarrow I=4{\int }_0^{\frac{\pi }{2}}{\cos }^{2}x{\sin }^{4}xdx$.......(2)

Add (1) and (2)

$2I=4{\int }_0^{\frac{\pi }{2}}{\cos }^{2}x{\sin }^{2}xdx$, since ${\sin }^{2}x+{\cos }^{2}x=1$

$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}(2\cos x\sin x{)}^{2}dx={\int }_0^{\frac{\pi }{2}}{\sin }^{2}2xdx={\int }_0^{\frac{\pi }{2}}\frac{1-\cos 4x}{2}dx$

$\Rightarrow 4I={\int }_0^{\frac{\pi }{2}}(1-\cos 4x)dx={\int }_0^{\frac{\pi }{2}}dx-{\int }_0^{\frac{\pi }{2}}\cos 4xdx$

$\Rightarrow 4I=\frac{\pi }{2}-0$

$\therefore I=\frac{\pi }{8}$