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Math Help - trig substitutions

  1. #1
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    trig substitutions

    I have the integration of radical(25 - x^2)/(x)
    i set x = 5sinx
    dx = 5cosx
    i ended up with integral of cos^2(x)/sin(x)
    ..not so sure where to go from here. Thanks for the help.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    I have the integration of radical(25 - x^2)/(x)
    i set x = 5sinx
    dx = 5cosx
    i ended up with integral of cos^2(x)/sin(x)
    ..not so sure where to go from here. Thanks for the help.
    \int \frac{\sqrt{25 - x^2}}{x} dx

    Let x = 5 sin(u) --> dx = 5 cos(u) du

    Thus
    \int \frac{\sqrt{25 - x^2}}{x} dx = \int \frac{\sqrt{25 - 25sin^2(u)}}{5sin(u)} 5 cos(u) du

    = 5 \sqrt{5} \int \frac{cos^2(u)}{sin(u)}du

    = 5 \sqrt{5} \int \frac{cos^2(u) - 1 + 1}{sin(u)}du

    = 5 \sqrt{5} \int \frac{cos^2(u) - 1}{sin(\theta)} du + 5 \sqrt{5} \int \frac{1}{sin(u)}du

    = -5 \sqrt{5} \int sin(\theta) du + 5 \sqrt{5} \int \frac{1}{sin(u)}du

    Is that any better?

    -Dan
    Last edited by topsquark; August 1st 2007 at 04:15 AM.
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  3. #3
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    Quote Originally Posted by topsquark View Post
    \int \frac{\sqrt{25 - x^2}}{x} dx

    Let x = 5 sin(u) --> dx = 5 cos(u) du

    Thus
    \int \frac{\sqrt{25 - x^2}}{x} dx = \int \frac{\sqrt{25 - 25sin^2(u)}}{\sqrt{5} \sqrt{sin(u)}} 5 cos(u) du

    = 5 \sqrt{5} \int \frac{cos^2(u)}{\sqrt{sin(u)}}du

    Can you take it from here?

    -Dan
    I dont really see what you can do with the intergral of cos^2(u)/rad(sin(u)
    the 2nd power in the cos makes it tough.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by davecs77 View Post
    I dont really see what you can do with the intergral of cos^2(u)/rad(sin(u)
    the 2nd power in the cos makes it tough.
    Ugh. I must have been tired last night. I have fixed my previous post.

    -Dan
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