Question

The value of the definite integral $\underset{0}{\overset{1/\sqrt{2}}{\int }}\frac{x^{2}dx}{\sqrt{1-x^2}\left(1+\sqrt{1-x^2}\right)}$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{4}+\frac{1}{\sqrt{2}}$
• C. $\frac{\pi }{4}-\frac{1}{\sqrt{2}}$
• D. none

Answer 1

Answer: (C)

$I=\underset{0}{\overset{1/\sqrt{2}}{\int }}\frac{x^{2}dx}{\sqrt{1-x^2}\left(1+\sqrt{1-x^2}\right)}$
put $x=\sin \theta ;dx=\cos \theta d\theta$
$=\underset{0}{\overset{\pi /4}{\int }}\frac{{\sin }^2\theta \cos \theta d\theta }{\cos \theta (1+\cos \theta )}=\underset{0}{\overset{\pi /4}{\int }}(1-\cos \theta )d\theta ==(\theta -\sin \theta {)}_0^{\pi /4}=\left(\frac{\pi }{4}-\frac{1}{\sqrt{2}}\right)$