<s>[INST] Definite integrals: Evaluate the integral 
Evaluate:
  $$ \int ^{1/2}_{1/4} \frac{dx}{ \sqrt{x-x^2}}dx$$

can u help me with this?
What is meant by the dx in the numerator?
EDIT:
ANSWER AS GIVEN IN THE BOOK
$$ \int ^{1/2}_{1/4} \frac{dx}{ \sqrt{x-x^2}}dx$$
$$=\int ^{1/2}_{1/4} \frac{1}{\sqrt {-(x^2-x+\frac{1}{4}- \frac{1}{4})}}dx$$
$$=\int ^{1/2}_{1/4} \frac{1}{\sqrt{({\frac{1}{2}})^2-(x-\frac{1}{2})^2}}$$
$$=\int ^{1/2}_{1/4} \frac{dx}{\sqrt{({\frac{1}{2}})^2-(x-\frac{1}{2})^2}}$$
$$= [ \arcsin\frac{x-1/2} {1/2} ]^{1/2}_{1/4}= \arcsin0-\arcsin\frac{-1}{2}$$
$$=\pi/6$$
Please offer your assistance :)
 [/INST] $$I=\int_{\frac14}^{\frac12}\frac{dx}{\sqrt{\left(\frac12\right)^2-\left(x-\frac12\right)^2}} =\int_{\frac14}^{\frac12}\frac{2dx}{\sqrt{1^2-(2x-1)^2}}$$
Put $2x-1=\sin y\implies 2dx=\cos y dy,$
When $x=\frac12,\sin y=0\implies y=0$ and $x=\frac14,\sin y=-\frac12\implies y=-\frac\pi6$
$$\implies I=\int_{-\frac\pi6}^0\frac{\cos ydy}{\cos y}=0-\left(-\frac\pi6\right)=\frac\pi6$$
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