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Thread: [SOLVED] trig substitution integral

  1. #1
    Senior Member mollymcf2009's Avatar
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    [SOLVED] trig substitution integral

    Evaluate the integral:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{dx}{x^3\sqrt{x^2-1}}$

    Here is what I am getting in my first step:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^3 \cdot (sec\theta - 1)} d\theta $

    Is that the right thing to do with the $\displaystyle x^3$ in the denominator?

    After that, assuming it is correct, I simplified the denominator:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^4 - (sec\theta)^3} d\theta$

    Is this correct to this point? If yes, where do I go from here?

    THanks!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Evaluate the integral:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{dx}{x^3\sqrt{x^2-1}}$

    Here is what I am getting in my first step:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^3 \cdot ({\color{red}sec\theta - 1})} d\theta $

    Is that the right thing to do with the $\displaystyle x^3$ in the denominator? Yes
    .
    .
    .
    what is in red is incorrect. can you fix it?
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  3. #3
    o_O
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    Also, change the limits of integration: $\displaystyle x = \sec \theta$

    • $\displaystyle \sqrt{2} = \sec \theta \ \Rightarrow \ \theta = \frac{\pi}{4}$
    • $\displaystyle {\color{white}.} \ 2 = \sec \theta \ \Rightarrow \ \theta = \frac{\pi}{3}$
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by Jhevon View Post
    what is in red is incorrect. can you fix it?
    [quote=mollymcf2009;264793]Evaluate the integral:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{dx}{x^3\sqrt{x^2-1}}$

    Here is what I am getting in my first step:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^3 \cdot (\sqrt{sec^2\theta - 1})} d\theta $

    Is that the right thing to do with the $\displaystyle x^3$ in the denominator?


    Better? What is my next step?
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  5. #5
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by o_O View Post
    Also, change the limits of integration: $\displaystyle x = \sec \theta$

    • $\displaystyle \sqrt{2} = \sec \theta \ \Rightarrow \ \theta = \frac{\pi}{4}$
    • $\displaystyle {\color{white}.} \ 2 = \sec \theta \ \Rightarrow \ \theta = \frac{\pi}{3}$
    But I need my answer in terms of x for webassign, should I still change them?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    [quote=mollymcf2009;264800]
    Quote Originally Posted by mollymcf2009 View Post
    Evaluate the integral:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{dx}{x^3\sqrt{x^2-1}}$

    Here is what I am getting in my first step:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^3 \cdot (\sqrt{sec^2\theta - 1})} d\theta $

    Is that the right thing to do with the $\displaystyle x^3$ in the denominator?


    Better? What is my next step?
    note that $\displaystyle \sec^2 \theta - 1 = \tan^2 \theta$

    with that in mind, simplify the integrand before you integrate

    Quote Originally Posted by mollymcf2009 View Post
    But I need my answer in terms of x for webassign, should I still change them?
    your answer will be a constant... either you use the x-limits with x or the u-limits with u. your choice. but you cannot write the x-limits on the u-integral
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  7. #7
    o_O
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    It's a definite integral. You'll be entering a numerical value. So all you need here is consistency.

    ________________

    $\displaystyle \int \frac{\sec \theta \tan \theta d \theta}{\sec^3 \theta \sqrt{\sec^2 \theta - 1}}$

    Cancel the $\displaystyle \sec \theta$ from the numerator with one in the denominator: $\displaystyle \int \frac{\tan \theta d \theta}{\sec^2 \theta\sqrt{\sec^2 \theta - 1}}$

    Use the identity $\displaystyle \sec^2 \theta - 1 = \tan^2 \theta$ to get: $\displaystyle \int \frac{\tan \theta d \theta}{\sec^2 \theta \cdot \sqrt{\tan^2 \theta}}$

    etc. etc.
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  8. #8
    Senior Member mollymcf2009's Avatar
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    Just want to double check before I continue:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{tan^2\theta \cdot sec\theta \cdot \sqrt{tan^2\theta}} d\theta$
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Just want to double check before I continue:

    $\displaystyle \int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{tan^2\theta \cdot sec\theta \cdot \sqrt{tan^2\theta}} d\theta$
    have you read post #7?
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  10. #10
    Senior Member mollymcf2009's Avatar
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    Sorry! I didn't see it until after I posted. I've got it now. Thanks everyone for your help
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  11. #11
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by o_O View Post
    It's a definite integral. You'll be entering a numerical value. So all you need here is consistency.

    ________________

    $\displaystyle \int \frac{\sec \theta \tan \theta d \theta}{\sec^3 \theta \sqrt{\sec^2 \theta - 1}}$

    Cancel the $\displaystyle \sec \theta$ from the numerator with one in the denominator: $\displaystyle \int \frac{\tan \theta d \theta}{\sec^2 \theta\sqrt{\sec^2 \theta - 1}}$

    Use the identity $\displaystyle \sec^2 \theta - 1 = \tan^2 \theta$ to get: $\displaystyle \int \frac{\tan \theta d \theta}{\sec^2 \theta \cdot \sqrt{\tan^2 \theta}}$

    etc. etc.
    This is driving me crazy! I have never had this much trouble grasping a concept.

    I'm gun shy about canceling for some reason. I feel like I need to do a u substitution instead. Can I cancel out $\displaystyle tan\theta$ in top and bottom since the bottom is sqrt of $\displaystyle tan^2$ ?? Leaving me with $\displaystyle \frac{1}{sec^2\theta}$ or am I supposed to use a u sub?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    This is driving me crazy! I have never had this much trouble grasping a concept.

    I'm gun shy about canceling for some reason. I feel like I need to do a u substitution instead. Can I cancel out $\displaystyle tan\theta$ in top and bottom since the bottom is sqrt of $\displaystyle tan^2$ ?? Leaving me with $\displaystyle \frac{1}{sec^2\theta}$
    yes, this is what you should do.

    now, note that $\displaystyle \frac 1{\sec^2 x} = \cos^2 x$
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  13. #13
    Senior Member mollymcf2009's Avatar
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    Jhevon! Thank you so much for your patience and help! I greatly appreciate it!! Molly
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