# Integration Query

• Jun 24th 2010, 07:02 PM
Lukybear
Integration Query
When I intergrated:

1/sqrt(x)sqrt(1-x) = I, Ive got sinh sqrtx. This i used a subsitution x=u^2

However, expanding the bottom;
I = 1/sqrt(x-x^2), then chaging the denominator to a perfect square:
I = 1/sqrt (1/4 - (1/2-x)^2), then using the same sin inverse identity, i got a complete different answer.

Can someone please explain to me this?
• Jun 24th 2010, 08:19 PM
simplependulum
Quote:

Originally Posted by Lukybear
When I intergrated:

1/sqrt(x)sqrt(1-x) = I, Ive got sinh sqrtx. This i used a subsitution x=u^2

However, expanding the bottom;
I = 1/sqrt(x-x^2), then chaging the denominator to a perfect square:
I = 1/sqrt (1/4 - (1/2-x)^2), then using the same sin inverse identity, i got a complete different answer.

Can someone please explain to me this?

So you have two different answers , one is $\displaystyle 2 \sin^{-1}(\sqrt{x}) + C$ and the other is $\displaystyle \sin^{-1}(2x-1) + K$ .

Consider $\displaystyle \cos(2t) = 1 - 2\sin^2(t)$

Also $\displaystyle \cos(2t) = \sin( \frac{\pi}{2} - 2t ) = 1 - 2\sin^2(t)$

Therefore , $\displaystyle \frac{\pi}{2} - 2t = \sin^{-1}(1 - 2\sin^2(t))$

Sub $\displaystyle \sin(t) = x$so we have $\displaystyle \frac{\pi}{2} - 2 \sin^{-1}(x) = \sin^{-1}(1 - 2x^2 ) = - \sin^{-1}(2x^2-1)$

$\displaystyle 2 \sin^{-1}(x) = \frac{\pi}{2} + \sin^{-1}(2x^2-1)$ or

$\displaystyle 2 \sin^{-1}(\sqrt{x}) = \frac{\pi}{2} + \sin^{-1}(2x-1)$

Then compare the two answers . Here $\displaystyle K = \frac{\pi}{2} + C$ .