Originally Posted by

**wutever** okk i know that but i have a weired ans :

1/4 sin inverse (2x) + 1/2 x sqrt (1-4x^2)

Mr F says: *Ahem* .... + C

so how ??

Mr F says: The answer you got is correct. If you differentiated it, you would see that you get \sqrt{1 - 4x^2} .......

and why $\displaystyle

\int \sqrt{1-4x^2}~dx

$

is equal to

$\displaystyle

\int \frac{1-4x^2}{\sqrt{1-4x^2}}~dx

$

Mr F says: Basic algebra. Multiply $\displaystyle {\color{red}\sqrt{1 - 4x^2}}$ by $\displaystyle {\color{red}1 = \frac{\sqrt{1 - 4x^2}}{\sqrt{1 - 4x^2}}}$