Question

Evaluate ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{2}x{\cos }^{4}xdx.$

Answer 1

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

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

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

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

$\Rightarrow 8I={\int }_0^{\frac{\pi }{2}}(1-\cos 4x)dx={\int }_0^{\frac{\pi }{2}}dx-{\int }_0^{\frac{\pi }{2}}\cos 4xdx$

$\Rightarrow 8I=\frac{\pi }{2}-0$

$\therefore I=\frac{\pi }{16}$