yet another trig substitution

**mollymcf2009** · February 12th 2009, 07:48 PM

Last one on this homework!!!

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

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.

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

Am I correct to this point?

Thanks!!

**mr fantastic** · February 12th 2009, 08:46 PM

Originally Posted by mollymcf2009

Last one on this homework!!!

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

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.

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

Am I correct to this point?

Thanks!!

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

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

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

**mollymcf2009** · February 12th 2009, 09:08 PM

Originally Posted by mr fantastic

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

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

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

I actually did do my substitution with

$x=10sec\theta$

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

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

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

Is that not correct?

**mr fantastic** · February 13th 2009, 02:29 AM

Originally Posted by mollymcf2009

Last one on this homework!!!

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

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.

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

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.

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