# Thread: How to integrate function?

1. ## How to integrate function?

Sorry guys, am very rusty in calculus.
What method should I use to evaluate this?

$\displaystyle \int{{1\over{\pi\sqrt{x(1-x)}}}}$

2. Originally Posted by chopet
Sorry guys, am very rusty in calculus.
What method should I use to evaluate this?

$\displaystyle \int{{1\over{\pi\sqrt{x(1-x)}}}}$
$\displaystyle \frac{1}{\sqrt{x(1-x)}}=\frac{1}{\sqrt{ x - x^2}} = \frac{1}{\sqrt{\frac{1}{4} - \left( x - \frac{1}{2} \right)^2}}$
Now let $\displaystyle t=x-1/2$ the rest is trivial.

3. Originally Posted by chopet
Sorry guys, am very rusty in calculus.
What method should I use to evaluate this?

$\displaystyle \int{{1\over{\pi\sqrt{x(1-x)}}}}$
Define $\displaystyle x=\frac1u,$ the integral becomes to

$\displaystyle - \int {\frac{1}{{u\sqrt {u - 1} }}\,du}.$

Make another substitution defined by $\displaystyle \varphi=\sqrt{u-1},$

$\displaystyle - \int {\frac{1} {{u\sqrt {u - 1} }}\,du} = - 2\int {\frac{1} {{\varphi ^2 + 1}}\,d\varphi } = - 2\arctan \varphi + k.$

Back substitute

$\displaystyle \int {\frac{1} {{\sqrt {x - x^2 } }}\,dx} = - 2\arctan \frac{{\sqrt {1 - x} }} {{\sqrt x }} + k.$

($\displaystyle \pi$ is just bothering the integrand.)