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Math Help - yet another trig substitution

  1. #1
    Senior Member mollymcf2009's Avatar
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    yet another trig substitution

    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!!
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  2. #2
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    Quote Originally Posted by mollymcf2009 View Post
    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}.
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by mr fantastic View Post
    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?
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  4. #4
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    Quote Originally Posted by mollymcf2009 View Post
    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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