substitution problem

• August 19th 2009, 12:03 PM
CarrieB
substitution problem
hey i've got a question i'm finding really tricky. if anyone can put this in latex i'd be really thankful!

use the substitution x-p = 1/u to show the integral

dx/(x - p)sqrt[(x - p)(x - q)] equals

2/(q - p) . sqrt[(x - q)/(x - p)]

thank you!
• August 19th 2009, 01:03 PM
red_dog
$I=\int\frac{dx}{(x-p)\sqrt{(x-p)(x-q)}}=\frac{2}{q-p}\sqrt{\frac{x-q}{x-p}}$

$x-p=\frac{1}{u}\Rightarrow dx=-\frac{1}{u^2}du$

$I=\int\frac{-\frac{1}{u^2}du}{\frac{1}{u}\sqrt{\frac{1}{u}\left (\frac{1}{u}+p-q\right)}}=$

$=-\int\frac{du}{\sqrt{1+u(p-q)}}=\frac{2}{q-p}\sqrt{1+u(p-q)}+C$

Now replace u with $\frac{1}{x-p}$.
• August 19th 2009, 01:05 PM
Mauritzvdworm
do you mean you need to show the following
$\int \frac{dx}{(x-p)\sqrt{(x-p)(x-q)}}=\frac{2}{p-q}\sqrt{\frac{x-q}{x-p}}$
using the substitution $x-p=\frac{1}{u}$??