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Math Help - Trig Sub Problems:

  1. #1
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    Trig Sub Problems:

    Hi guys. I have two problems that revolve around trig sub and just want to make sure that I got these right.

    1) integral of cos (sqrt(x))^2 / (sqrt(x))

    I got (sqrt) x + (1/2)sin(2(sqrt(x)) + C.

    2) integral of (x^3) / sqrt (x^2 + 4), with limits from 0 to 2.

    For this one, I'm a little confused on how to set this one up. Which trig identity do I use that will successfully satisfy this beginning?
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  2. #2
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    Quote Originally Posted by XFactah416 View Post
    Hi guys. I have two problems that revolve around trig sub and just want to make sure that I got these right.

    1) integral of cos (sqrt(x))^2 / (sqrt(x))

    I got (sqrt) x + (1/2)sin(2(sqrt(x)) + C.

    2) integral of (x^3) / sqrt (x^2 + 4), with limits from 0 to 2.

    For this one, I'm a little confused on how to set this one up. Which trig identity do I use that will successfully satisfy this beginning?
    Personally i would not use trig sub for either of these. a substitution of u = \sqrt x works nicely for the first (is the square on the cosine? if so, you are correct) and a substitution of u^2 = x^2 + 4 works wonders for the second. if you insist on doing trig sub for this, you would use x = 2 \tan \theta, the first does not have a required form to apply trig sub.
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    Quote Originally Posted by Jhevon View Post
    Personally i would not use trig sub for either of these. a substitution of u = \sqrt x works nicely for the first (is the square on the cosine? if so, you are correct) and a substitution of u^2 = x^2 + 4 works wonders for the second. if you insist on doing trig sub for this, you would use x = 2 \tan \theta, the first does not have a required form to apply trig sub.
    Yeah, the square is on the cosine for the first one.

    Thanks a lot!
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    Sorry, hate to bring this topic back up, but I would like to know the answer to #2.

    It seems like after I do a substitiution of x = 2 tan (theta), my limits of integration go to 0 to -4.37. Then, after some algebra, and another substitution (this time of u = sec (theta), my limits of integration is now 1 to -2.98.
    Last edited by XFactah416; February 15th 2010 at 08:54 PM.
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    Quote Originally Posted by XFactah416 View Post
    Sorry, hate to bring this topic back up, but I would like to know the answer to #2.

    It seems like after I do a substitiution of x = 2 tan (theta), my limits of integration go to 0 to -4.37. Then, after some algebra, and another substitution (this time of u = sec (theta), my limits of integration is now 1 to -2.98.
    \int_0^2\frac{x^3}{\sqrt{x^2+4}}\,dx

    =\int_0^{\frac{\pi}{4}}\frac{8tan^3{\theta}}{2sec{  \theta}}2sec^2{\theta}\,d\theta

    =\int_0^{\frac{\pi}{4}}8tan^2{\theta}sec{\theta}ta  n{\theta}\,d\theta

    =\int_0^{\frac{\pi}{4}}8(sec^2{\theta}-1)sec{\theta}tan{\theta}\,d\theta

    =8\left[\frac{sec^3{\theta}}{3}-sec{\theta}\right]_0^{\frac{\pi}{4}}
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  6. #6
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    Quote Originally Posted by XFactah416 View Post
    2) integral of (x^3) / sqrt (x^2 + 4), with limits from 0 to 2.
    Hyperbolic Substitution is better for this integral.

    Let x = 2\sinh{t} so that dx = 2\cosh{t}\,dt.


    So the integral becomes

    \int{\frac{x^3}{\sqrt{x^2 + 4}}\,dx} = \int{\frac{(2\sinh{t})^3}{\sqrt{(2\sinh{t})^2 + 4}}\,2\cosh{t}\,dt}

     = \int{\frac{16\sinh^3{t}\cosh{t}}{\sqrt{4\sinh^2{t} + 4}}\,dt}

     = \int{\frac{16\sinh^3{t}\cosh{t}}{\sqrt{4(\sinh^2{t  } + 1)}}\,dt}

     = \int{\frac{16\sinh^3{t}\cosh{t}}{2\sqrt{\cosh^2{t}  }}\,dt}

     = \int{\frac{8\sinh^3{t}\cosh{t}}{\cosh{t}}\,dt}

     = 8\int{\sinh^3{t}\,dt}

     = 8\int{\sinh{t}\sinh^2{t}\,dt}

     = 8\int{\sinh{t}(\cosh^2{t} - 1)\,dt}


    Now let u = \cosh{t} so that \frac{du}{dt} = \sinh{t}, so that the integral becomes

     = 8\int{(u^2 - 1)\,\frac{du}{dt}\,dt}

     = 8\int{u^2 - 1\,du}

     = 8\left[\frac{1}{3}u^3 - u\right] + C

     = 8\left[\frac{1}{3}\cosh^3{t} - \cosh{t}\right] + C

     = \frac{8}{3}\cosh^3{t} - 8\cosh{t} + C.


    And since x = 2\sinh{t}

    \frac{x}{2} = \sinh{t}

    \frac{x^2}{4} = \sinh^2{t}

    \frac{x^2}{4} = \cosh^2{t} - 1

    1 + \frac{x^2}{4} = \cosh^2{t}

    \cosh{t} = \sqrt{1 + \frac{x^2}{4}}.


    So finally, your integral is...

    \frac{8}{3}\sqrt{\left(1 + \frac{x^2}{4}\right)^2} - 8\sqrt{1 + \frac{x^2}{4}} + C.
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