Question

If $P={\int }_0^1{x}^{n-1}(1-x{)}^{n}dx$ and $l_{2n-1}={\int }_0^{\pi }(\sin x{)}^{2n-1}dx$ then $\frac{l_{2n-1}}{P}$ is

• A. $2^{2n}$
• B. $2^{2n-1}$
• C. $2^{2n-2}$
• D. None of these

Answer 1

Answer: (A)

$2^{2n}$

$P={\int }_0^1{X}^{n-1}(1-x{)}^{n}dx$ ............(1)
$X\to (1-x)P={\int }_0^1(1-x{)}^{n-1}{X}^{n}dx$ ..............(2)
Equation (1) + (2)
$2P={\int }_0^1{X}^{n-1}(1-x{)}^{n-1}(1-x+x)dx\Rightarrow 2P{\int }_0^1{X}^{n-1}(1-x{)}^{n-1}dx$
Let $x={\sin }^2\theta ,dx=2\sin \theta \cos \theta d\theta$
$2P={\int }_0^{\pi /2}\left(\sin \theta {)}^{2n-2}\right(\cos \theta {)}^{2n}-2.\sin 2\theta d\theta \Rightarrow 2P=\frac{1}{2^{2n-2}}{\int }_0^{\pi /2}{2}^{2n-2}\left(\sin \theta {)}^{2n-2}\right(\cos \theta {)}^{2n-2}.\sin 2\theta d\theta$
$2P=\frac{1}{2^{2n-2}}{\int }_0^{\pi /2}(\sin 2\theta {)}^{2n-2}.\sin 2\theta d\theta \Rightarrow 2P=\frac{1}{2^{2n-2}}{\int }_0^{\pi /2}(\sin 2\theta {)}^{2n-1}d\theta$
Let $2\theta =t,d\theta =\frac{dt}{2}\Rightarrow 2P=\frac{1}{2^{2n-1}}{\int }_0^{\pi }(\sin t{)}^{2n-1}dt=\frac{1}{2^{2n-1}}.l_{2n-1}$
$2^{2n}=\frac{l_{2n-1}}{P}$