Question

Obtain the reduction formula for ${\int }_0^{\pi /2}{\sin }^{n}xdx$ for an integer $n\ge 2$.

Answer 1

Let $I_n={\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx={\int }_0^{\frac{\pi }{2}}{\sin }^{n-1}xsinxdx$

now do integration by parts , we get $I_n={\sin }^{n-1}x(-cosx)+{\int }_0^{\frac{\pi }{2}}(n-1){\cos }^{2}x{\sin }^{n-2}x=0+(n-1){\int }_0^{\frac{\pi }{2}}{\sin }^{n-2}x(1-{\sin }^{2}x)$

$=(n-1)I_{n-2}-(n-1)I_n$

$\Rightarrow n{I}_n=(n-1)I_{n-2}$