Question

If $I={\int }_0^{\pi /2}{\cos }^{n}x{\sin }^{n}xdx=\lambda {\int }_0^{\pi /2}{\sin }^{n}xdx$ then $\lambda$ equals

• A. $2^{-n+1}$
• B. $2^{-n-1}$
• C. $2^{-n}$
• D. $2^{-1}$

Answer 1

The correct option is C $2^{-n}$

$I=\frac{1}{2^n}{\int }_0^{\pi /2}(2\sin x\cos x{)}^{n}dx$
$=\frac{1}{2^n}{\int }_0^{\pi /2}(\sin 2x{)}^{n}dx$
Substitute $2x=\theta ,$ so that
$I=\frac{1}{2^n}{\int }_0^{\pi }\left({\sin }^n\theta \right)\frac{1}{2}d\theta$
$=\frac{1}{2^{n+1}}\left[{\int }_0^{\pi /2}\right[(\sin \theta {)}^n+(\sin (\pi -\theta ){)}^n]d\theta$
Using ${\int }_0^{2a}f\left(x\right)dx,={\int }_0^a\left[f\right(x)+f(2a-x\left)\right]dx$
Thus,$I=\frac{1}{2^n}{\int }_0^{\pi /2}{\sin }^n\theta d\theta$

$\therefore \lambda =2^{-n}$