Question

Evaluate the integral
${\int }_0^1{\sin }^{-1}\sqrt{x}dx$

• A. $\pi$
• B. $-\pi$
• C. $\pi /4$
• D. $\pi /3$

Answer 1

Answer: (C)

$\pi /4$

${\int }_0^1{\sin }^{-1}\left(\sqrt{x}\right)dx$

Let $\sqrt{x}=t\Rightarrow x=t^2$

$dx=2tdt$

$\therefore {\int }_0^1{\sin }^{-1}\left(t\right)\cdot 2tdt$

$=2{\int }_0^1{\sin }^{-1}\left(t\right)\cdot t\cdot dt$

$=2{\left[{\sin }^{-1}\right(t)\cdot \frac{t^2}{2}]}_0^1–2\cdot {\int }_0^1\frac{\frac{t^2}{2}}{\sqrt{1–t^2}}dt$

$=\frac{\pi }{2}+{\int }_0^1=\frac{-t^2}{\sqrt{1–t^2}}dt$

Let $t=\sin \theta$

$=\frac{\pi }{2}+{\int }_0^{\pi /2}-\frac{{\sin }^2\theta }{\cos \theta }\cdot \cos \theta \cdot d\theta$

$=\frac{\pi }{2}-{\int }_0^{\pi /2}{\sin }^2\theta \cdot d\theta$

$=\frac{\pi }{2}-(\frac{1}{2}\times \frac{\pi }{2})$ (using walli’s formula)

$=\frac{\pi }{4}$

$\therefore {\int }_0^1{\sin }^{-1}\left(\sqrt{x}\right)dx=\frac{\pi }{4}$.