Question

Solve ${\int }_0^1\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$ is equal to

• A. $\pi /2$
• B. $\pi /4$
• C. $0$
• D. $1$

Answer 1

Answer: (D)

$1$

$I={\int }_0^1\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}dx$

Putting ${\sin }^{-1}x=t$

$\frac{1}{\sqrt{1-x^2}}dx=dt$

Changing limts $x:0\to 1::t;0\to \frac{\pi }{2}$

Thus $I={\int }_0^1\sin t\times t\times \frac{dx}{\sqrt{1-x^2}}={\int }_0^{\frac{\pi }{2}}\sin t\times tdt$

${\int }_0^{\frac{\pi }{2}}\sin t\times tdt=t{\int }_0^{\frac{\pi }{2}}\sin tdt-{\int }_0^{\frac{\pi }{2}}(\frac{d\left(t\right)}{dt}\int \sin tdt)dt$

$=\left[t\right(-\cos t){]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}(-\cos t)$

$=[-t(\cos t){]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}(\cos t)$

$=[-t(\cos t){]}_0^{\frac{\pi }{2}}+[\sin t{]}_0^{\frac{\pi }{2}}$

$=-\frac{\pi }{2}\cos \frac{\pi }{2}+\sin \frac{\pi }{2}-[-0\times \cos 0+\sin 0]$

$=-\frac{\pi }{2}\times 0+1+0\times 1=0$

$=1$