Question

If ${\int }_{1/4}^x\frac{dt}{\sqrt{t-t^2}}=\frac{\pi }{6}$, then $x$ equals

• A. $1/2$
• B. $1/3$
• C. $1$
• D. none of these

Answer 1

The correct option is A $1/2$

${\int }_{\frac{1}{4}}^x\frac{1}{\sqrt{t-t^2}}dt=\frac{\pi }{6}\Rightarrow {\int }_{\frac{1}{4}}^x\frac{1}{\sqrt{\frac{1}{4}-{\left(t\frac{1}{2}\right)}^2}}dt=\frac{\pi }{6}$
Substitute $t-\frac{1}{2}=u\Rightarrow dt=du$
$\therefore {\int }_{-\frac{1}{4}}^{x\frac{-1}{2}}\frac{1}{\sqrt{\frac{1}{4}-u^2}}du=\frac{\pi }{6}\Rightarrow 2{\int }_{\frac{-1}{4}}^{x\frac{-1}{2}}\frac{1}{\sqrt{1-{4u}^2}}=\frac{\pi }{6}$
$\Rightarrow {\left[si{n}^{-1}2t\right]}_{\frac{-1}{4}}^{x\frac{-1}{2}}=\frac{\pi }{6}\Rightarrow x=\frac{1}{2}$