Question

Evaluate: ${\int }_{1/2}^1\frac{1}{\sqrt{x-x^2}}dx$

• A. $\pi /8$
• B. $\pi /4$
• C. $\pi /2$
• D. $\pi$

Answer 1

The correct option is C $\pi /2$

${\int }_{1/2}^1\frac{1}{\sqrt{x-x^2}}dx$

$={\int }_{1/2}^1\frac{1}{\sqrt{{\left(\frac{1}{2}\right)}^2-{(x-\frac{1}{2})}^2}}dx$

we know that $\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\frac{x}{a}$

Now, our integration becomes:-

${\left[{\sin }^{-1}\left(\frac{x-\frac{1}{2}}{\frac{1}{2}}\right)\right]}_{1/2}^1$

$=[{\sin }^{-1}1-{\sin }^{-1}0]$

$=\frac{\pi }{2}$

Hence,answer is option_(C).