Question

The value of ${\int }_0^a\frac{1}{\sqrt{\mathrm{ax}-x^2}}\mathrm{dx}$ is

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\mathrm{a\pi }$
• D. $\frac{\pi }{a}$

Answer 1

The correct option is A $\pi$

We have,
${\int }_0^a\frac{1}{\sqrt{\mathrm{ax}-x^2}}\mathrm{dx}={\int }_0^a\frac{1}{\sqrt{-\left(x^2-\mathrm{ax}+\frac{a^2}{4}-\frac{a^2}{4}\right)}}\mathrm{dx}={\int }_0^a\frac{1}{\sqrt{{\left(\frac{a}{2}\right)}^2-{\left(x-\frac{a}{2}\right)}^2}}\mathrm{dx}={\left({\sin }^{-1}\left(\frac{x-\frac{a}{2}}{\frac{a}{2}}\right)\right]}_0^a={\left({\sin }^{-1}\left(\frac{2x-a}{a}\right)\right]}_0^a={\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(-1\right)=2{\sin }^{-1}\left(1\right)=2\left(\frac{\pi }{2}\right)=\pi$