# Integral- trig substitution?

• Feb 23rd 2010, 07:00 AM
daberkow
Integral- trig substitution?
Hi. I'm trying to solve this integral. I think I may need a trig substitution, but I've tried a ton and I'm lost. Any specific help would be much appreciated.

$\displaystyle \int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx$

The solution is

$\displaystyle \frac{x}{z^2(x^2+z^2)^{1/2}}$

• Feb 23rd 2010, 07:03 AM
Ted
Quote:

Originally Posted by daberkow
Hi. I'm trying to solve this integral. I think I may need a trig substitution, but I've tried a ton and I'm lost. Any specific help would be much appreciated.

$\displaystyle \int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx$

The solution is

$\displaystyle \frac{x}{z^2(x^2+z^2)^{1/2}}$

Assuming $\displaystyle z$ is a constant:
Yes. Triginometric Substitution. Let $\displaystyle x=ztan(\theta)$ then $\displaystyle dx=zsec^2(\theta)d\theta$.
Post your work; to see why did you stuck.
• Feb 23rd 2010, 02:07 PM
daberkow
$\displaystyle \int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx$

$\displaystyle \int\frac{zsec^2\theta}{(z^2+z^2tan^2\theta)^\frac {3}{2}}d\theta$

$\displaystyle \int\frac{zsec^2\theta}{((z^2(1+tan^2\theta))^\fra c{3}{2}}d\theta$

$\displaystyle \int\frac{zsec^2\theta}{(z^2(sec^2\theta))^\frac{3 }{2}}d\theta$

$\displaystyle \int\frac{sec^2\theta}{z^2(sec^2\theta))^\frac{3}{ 2}}d\theta$

$\displaystyle \int\frac{1}{z^2(sec^2\theta)^\frac{1}{2}}d\theta$

$\displaystyle \int\frac{1}{z^2(sec\theta)}d\theta$

I went wrong somewhere...
• Feb 23rd 2010, 02:33 PM
ione
Quote:

Originally Posted by daberkow
$\displaystyle \int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx$

$\displaystyle \int\frac{zsec^2\theta}{(z^2+z^2tan^2\theta)^\frac {3}{2}}d\theta$

$\displaystyle \int\frac{zsec^2\theta}{((z^2(1+tan^2\theta))^\fra c{3}{2}}d\theta$

$\displaystyle \int\frac{zsec^2\theta}{(z^2(sec^2\theta))^\frac{3 }{2}}d\theta$

$\displaystyle \int\frac{sec^2\theta}{z^2(sec^2\theta))^\frac{3}{ 2}}d\theta$

$\displaystyle \int\frac{1}{z^2(sec^2\theta)^\frac{1}{2}}d\theta$

$\displaystyle \int\frac{1}{z^2(sec\theta)}d\theta$

I went wrong somewhere...

Try using $\displaystyle \frac{1}{\sec\theta}=\cos\theta$
• Feb 23rd 2010, 02:35 PM
tom@ballooncalculus
Quote:

Originally Posted by daberkow
I went wrong somewhere...

No, keep going...

http://www.mathhelpforum.com/math-he...tml#post457096
• Feb 23rd 2010, 02:49 PM
Ted
Quote:

Originally Posted by daberkow
$\displaystyle \int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx$

$\displaystyle \int\frac{zsec^2\theta}{(z^2+z^2tan^2\theta)^\frac {3}{2}}d\theta$

$\displaystyle \int\frac{zsec^2\theta}{((z^2(1+tan^2\theta))^\fra c{3}{2}}d\theta$

$\displaystyle \int\frac{zsec^2\theta}{(z^2(sec^2\theta))^\frac{3 }{2}}d\theta$

$\displaystyle \int\frac{sec^2\theta}{z^2(sec^2\theta))^\frac{3}{ 2}}d\theta$

$\displaystyle \int\frac{1}{z^2(sec^2\theta)^\frac{1}{2}}d\theta$

$\displaystyle \int\frac{1}{z^2(sec\theta)}d\theta$

I went wrong somewhere...

Good. No mistakes.
You almost done.
$\displaystyle \int\frac{1}{z^2(sec\theta)}d\theta=\frac{1}{z^2}\ int cos(\theta) d\theta$