1. Trig Substitution

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

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

$\displaystyle x=a tan\theta$
$\displaystyle dx=sec^2\theta$

$\displaystyle \sqrt{(x^2+a^2)}= \sqrt{a^2tan^2\theta+a^2}= asec\theta\ ?$

$\displaystyle \int \frac {1}{({asec\theta})^3}sec^2\theta\ ?$

2. Looks good to me.

We must remember that $\displaystyle \sqrt{\sec^2\theta}=|\sec\theta|$, but since $\displaystyle \theta=\arctan\frac{x}{a}$, $\displaystyle \sec\theta$ is always positive.

3. Originally Posted by Scott H
Looks good to me.

We must remember that $\displaystyle \sqrt{\sec^2\theta}=|\sec\theta|$, but since $\displaystyle \theta=\arctan\frac{x}{a}$, $\displaystyle \sec\theta$ is always positive.
Ok thanks, I'm not sure what to do next though. u substitution? trig identities?

$\displaystyle \int \frac {1}{({asec\theta})^3}sec^2\theta=\$

4. Originally Posted by JJ007
Ok thanks, I'm not sure what to do next though. u substitution? trig identities?

$\displaystyle \int \frac {1}{({asec\theta})^3}sec^2\theta=\$

$\displaystyle \frac{1}{a^3} \int \cos{\theta}d\theta$

5. Originally Posted by 11rdc11
$\displaystyle \frac{1}{a^3} \int \cos{\theta}d\theta$
$\displaystyle \frac{sin{\theta\ d\theta } }{a^3}+c$
So is this the full answer to the problem or does it involve a triangle at some point?

6. Originally Posted by JJ007
$\displaystyle \frac{sin{\theta\ d\theta } }{a^3}+c$
So is this the full answer to the problem or does it involve a triangle at some point?
One more step, back sub for your the trig sub

$\displaystyle \frac{x}{a^3\sqrt{a^2+x^2}}+C$