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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 :)

chndn
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2 Answers2

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Let's assume the doubled $dx$ is a typo. For the integration, let $x=u^2$. Then $dx=2u\,du$. The bottom becomes $u\sqrt{1-u^2}$, so our integral is $$\int_{1/2}^{1/\sqrt{2} }\frac{2\,du}{\sqrt{1-u^2}}.$$ The integration is easy, we get an arcsine, and end up with $\frac{\pi}{6}$.

Remark: If we take the doubled $dx$ to be not a typo, but a deliberate attempt to confuse (not nice!) things are not much harder. Make the same substitution, and then let $u=\sin t$.

André Nicolas
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  • can u explain the remark. Assume that it isnt A TYPO – chndn Oct 06 '13 at 05:34
  • Forget about the $d$, it is a constant. When we do the substitution, we need to integrate $\frac{u^2}{\sqrt{1-u^2}}$. Let $u=\sin t$. Then $du=\cos t,dt$ and we end up with the integral of $2\sin^2 t,dt$. This is standard, for example using $\cos 2t=1-2\sin^2 t$. – André Nicolas Oct 06 '13 at 05:38
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$$I=\int_{\frac14}^{\frac12}\frac{dx}{\sqrt{\left(\frac12\right)^2-\left(x-\frac12\right)^2}} =\int_{\frac14}^{\frac12}\frac{2dx}{\sqrt{1^2-(2x-1)^2}}$$

Put $2x-1=\sin y\implies 2dx=\cos y dy,$

When $x=\frac12,\sin y=0\implies y=0$ and $x=\frac14,\sin y=-\frac12\implies y=-\frac\pi6$

$$\implies I=\int_{-\frac\pi6}^0\frac{\cos ydy}{\cos y}=0-\left(-\frac\pi6\right)=\frac\pi6$$

lab bhattacharjee
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