# If the value of the integral

Question:

If the value of the integral $\int_{0}^{1 / 2} \frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} d x$ is $\frac{k}{6}$, then

$k$ is equal to :

1. (1) $2 \sqrt{3}-\pi$

2. (2) $2 \sqrt{3}+\pi$

3. (3) $3 \sqrt{2}+\pi$

4. (4) $3 \sqrt{2}-\pi$

Correct Option: 1

Solution:

$\frac{k}{6}=\int_{0}^{\frac{1}{2}} \frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} d x$

Let $x=\sin \theta ; d x=\cos \theta d \theta$,

then $\int_{0}^{\frac{1}{2}} \frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} d x=\int_{0}^{\frac{\pi}{6}} \frac{\sin ^{2} \theta \cos \theta}{\cos ^{3} \theta} d \theta$

$\therefore \frac{k}{6}=\int_{0}^{\frac{\pi}{6}} \frac{\sin ^{2} \theta}{\cos ^{3} \theta} \cdot \cos \theta d \theta$

$\Rightarrow \frac{k}{6}=\int_{0}^{\frac{\pi}{6}} \tan ^{2} \theta d \theta=\int_{0}^{\frac{\pi}{6}}\left(\sec ^{2} \theta-1\right) d \theta$

$\Rightarrow \frac{k}{6}=(\tan \theta-\theta)_{0}^{\pi / 6}=\left(\frac{1}{\sqrt{3}}-\frac{\pi}{6}\right)=\frac{2 \sqrt{3}-x}{6}$

$\Rightarrow k=2 \sqrt{3}-\pi$