# Tough integral

1. Oct 24, 2010

### Zorba

I am trying to figure out which substitution to use to get this integral done:

$$\int \frac{du}{\sqrt{u-u^2} \cdot (1+ub)}$$

When I plug it into Mathematica I get:

$$\sqrt{\frac{4}{b+1}} \cdot \texttt{arctan} \left ( \sqrt{\frac{(b+1)u}{1-u}} \right )$$

Any ideas about a suitable substitution?

2. Oct 24, 2010

### HallsofIvy

I would complete the square in the square root: $u- u^2= -(u^2- u+ 1/4- 1/4)= 1/4- (u- 1/2)^2$.

Then let v= u- 1/2 so that u= v+ 1/2, 1+ ub= (1+ (1/2)b)+ bv.

Now the integral is
$$\int\frac{dv}{\sqrt{1/4- v^2}((1+ (1/2)b)+ bv)}$$
and $v= (1/2)sin(\theta)$ looks like a plausible substitution.