Question

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}x\sqrt{1-x^2}dx=$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{8}$

Answer 1

The correct option is B $\frac{\pi }{3}$

${\int }_0^{\frac{\pi }{2}}x\sqrt{1-x^2}dx$

Let $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$

$={\int }_0^{\frac{\pi }{2}}\sin \theta \sqrt{1-{\sin }^2\theta }\cos \theta d\theta$

$={\int }_0^{\frac{\pi }{2}}\sin \theta \sqrt{{\cos }^2\theta }\cos \theta d\theta$

$={\int }_0^{\frac{\pi }{2}}\sin \theta {\cos }^2\theta d\theta$

$={\int }_0^{\frac{\pi }{2}}\sin \theta \cos \theta \cos \theta d\theta$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}2\sin \theta \cos \theta \cos \theta d\theta$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\sin 2\theta \cos \theta d\theta$

$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}2\sin 2\theta \cos \theta d\theta$

$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}(\sin 3\theta +\sin \theta )d\theta$

$=\frac{1}{4}{[\frac{-\cos 3\theta }{3}-\cos \theta ]}_0^{\frac{\pi }{2}}$

$=\frac{-1}{4}{[\frac{\cos 3\theta }{3}+\cos \theta ]}_0^{\frac{\pi }{2}}$

$=\frac{-1}{4}[\frac{\cos 3\frac{\pi }{2}}{3}-\frac{\cos 0}{3}+\cos \frac{\pi }{2}-\cos 0]$

$=-\frac{\pi }{4}[0-\frac{1}{3}+0-1]$

$=-\frac{\pi }{4}[-\frac{1+3}{3}]$

$=\frac{\pi }{4}\left[\frac{4}{3}\right]=\frac{\pi }{3}$