1. ## Evaluating Irrational Integral

I need help to integrate this particular integral step-by-step. I've been wrestling around with substitution method without any satisfying result. I don't know I'm doing wrong with it.

$\displaystyle \int_{-a}^a \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy$

This is my attempt:
Let u = $\displaystyle \left(x^2+y^2\right)$
du = $\displaystyle 2y\text{dy}$

Hence:
$\displaystyle \int \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy \int \frac{u^{-3/2}}{2} \, du \frac{1}{2} \int u^{-3/2} \, du \frac{\frac{1}{\sqrt{u}}}{\frac{2 (-1)}{2}}-\frac{1}{\sqrt{u}}-\frac{1}{\sqrt{x^2+y^2}}-\frac{1}{\sqrt{x^2+y^2}}$
$\displaystyle -\frac{1}{\sqrt{x^2+y^2}}$

Thank You

2. ## Re: Evaluating Irrational Integral

Originally Posted by Lites
I need help to integrate this particular integral step-by-step. I've been wrestling around with substitution method without any satisfying result. I don't know I'm doing wrong with it.

$\displaystyle \int_{-a}^a \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy$

This is my attempt:
Let u = $\displaystyle \left(x^2+y^2\right)$
du = $\displaystyle 2y\text{dy}$

Hence:
$\displaystyle \int \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy \int \frac{u^{-3/2}}{2} \, du \frac{1}{2} \int u^{-3/2} \, du \frac{\frac{1}{\sqrt{u}}}{\frac{2 (-1)}{2}}-\frac{1}{\sqrt{u}}-\frac{1}{\sqrt{x^2+y^2}}-\frac{1}{\sqrt{x^2+y^2}}$
$\displaystyle -\frac{1}{\sqrt{x^2+y^2}}$

Thank You
note that since $\displaystyle du = 2y \, dy$ , $\displaystyle dy = \frac{du}{2y}$ ... you left out the $\displaystyle y$ in the denominator in your substitution.

try a trig substitution, maybe?

$\displaystyle y = x\tan{t}$