Question

Evaluate ${\int }_0^1\frac{{\sin }^{-1}\sqrt{x}}{x^2-x+1}dx$

• A. $\frac{{\pi }^2}{6\sqrt{3}}$
• B. $\frac{{\pi }^4}{3\sqrt{3}}$
• C. $\frac{{\pi }^4}{6\sqrt{3}}$
• D. $\frac{{\pi }^2}{3\sqrt{3}}$

Answer 1

The correct option is A $\frac{{\pi }^2}{6\sqrt{3}}$

$l={\int }_0^1\frac{{\sin }^{-1}\sqrt{x}}{x^2-x+1}dx$....(i)
$l={\int }_0^1\frac{{\sin }^{-1}\sqrt{1-x}}{{(1-x)}^2-(1-x)+1}dx={\int }_0^1\frac{{\sin }^{-1}\sqrt{1-x}}{x^2-x+1}dx$...(ii)
Add (i) and (ii)
$2l=\frac{\pi }{2}{\int }_0^1\frac{dx}{{(x-1/2)}^2+{\left(\sqrt{3}/2\right)}^2}$
$\Rightarrow l=\frac{\pi }{4}\frac{2}{\sqrt{3}}{\left[{\tan }^{-1}\frac{2x-1}{\sqrt{3}}\right]}_0^1=\frac{{\pi }^2}{6\sqrt{3}}$