_ 1 / 4 ^ 1 / 2 d x x-x^2 = - Vidyadip

MCQ

$\int_{1 / 4}^{1 / 2} \frac{d x}{\sqrt{x-x^2}}=$

A
$\pi$
B
$\frac{\pi}{2}$
C
$\frac{\pi}{3}$
✓
$\frac{\pi}{6}$

✓

Answer

Correct option: D.

$\frac{\pi}{6}$

(D)
$\int_{1 / 4}^{1 / 2} \frac{d x}{\sqrt{x-x^2}}=\int_{1 / 4}^{1 / 2} \frac{d x}{\sqrt{\left(\frac{1}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}}$
$=\left[\sin ^{-1}\left(\frac{x-\frac{1}{2}}{\frac{1}{2}}\right)\right]_{\frac{1}{4}}^{\frac{1}{2}}$
$=\left[\sin ^{-1}(2 x-1)\right]_{1 / 4}^{1 / 2}=\frac{\pi}{6}$

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