# yet another trig substitution

• Feb 12th 2009, 07:48 PM
mollymcf2009
yet another trig substitution
Last one on this homework!!!

$\displaystyle \int \frac{\sqrt{x^2-100}}{x^3}dx$

Here is where I have gotten to:
*I left out the first 10 steps I did, let me know if you wanna see all those.

$\displaystyle \frac{1}{100} \int \frac{tan^2\theta}{sec^2\theta} d\theta$

Am I correct to this point?

Thanks!!
• Feb 12th 2009, 08:46 PM
mr fantastic
Quote:

Originally Posted by mollymcf2009
Last one on this homework!!!

$\displaystyle \int \frac{\sqrt{x^2-100}}{x^3}dx$

Here is where I have gotten to:
*I left out the first 10 steps I did, let me know if you wanna see all those.

$\displaystyle \frac{1}{100} \int \frac{tan^2\theta}{sec^2\theta} d\theta$

Am I correct to this point?

Thanks!!

No. It looks like you made the substitution $\displaystyle x = 10 \tan \theta$ and then applied some incorrect identities.

You need to make the substitution $\displaystyle x = 10 \sec \theta \Rightarrow dx = \frac{10 \sin \theta}{\cos^2 \theta} \, d \theta$.

Note that $\displaystyle 1 + \tan^2 \theta = \frac{1}{\cos^2 \theta}$.
• Feb 12th 2009, 09:08 PM
mollymcf2009
Quote:

Originally Posted by mr fantastic
No. It looks like you made the substitution $\displaystyle x = 10 \tan \theta$ and then applied some incorrect identities.

You need to make the substitution $\displaystyle x = 10 \sec \theta \Rightarrow dx = \frac{10 \sin \theta}{\cos^2 \theta} \, d \theta$.

Note that $\displaystyle 1 + \tan^2 \theta = \frac{1}{\cos^2 \theta}$.

I actually did do my substitution with

$\displaystyle x=10sec\theta$

$\displaystyle dx = 10sec\theta tan\theta d\theta$

Which looks like what you did but I didn't change it to sin & cos. I took my $\displaystyle 10sec\theta tan\theta d\theta$ and it got rid of the one in my problem leaving me with:

$\displaystyle \int \frac{\sqrt{100u^2 - 100}}{1000u^3}$

Is that not correct?
• Feb 13th 2009, 02:29 AM
mr fantastic
Quote:

Originally Posted by mollymcf2009
Last one on this homework!!!

$\displaystyle \int \frac{\sqrt{x^2-100}}{x^3}dx$

Here is where I have gotten to:
*I left out the first 10 steps I did, let me know if you wanna see all those.

$\displaystyle \frac{1}{100} \int \frac{tan^2\theta}{sec^2\theta} d\theta$

Am I correct to this point?

Thanks!!

OK, I should have looked more carefully ....

Nevertheless, I think you'll find the factor should be 1/10, not 1/100.