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Math Help - Integration with Trig. Substitution

  1. #1
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    Integration with Trig. Substitution

    Hey guys:
    Had one problem with my math homework:

    \int \frac{y^2}{25+y^2} dy

    So I tried y=5\tan x

    So then: dy=5\sec^2xdx,

    So the integration appears as: \int \frac{25\tan^2x}{25(1+\tan^2x)}(5\sec^2x)dx

    Which simplifies to: \int 5\tan^2x dx, which is where I get stuck.

    Any advice on moving forward, or a different, easier method?

    Peter
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  2. #2
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    Quote Originally Posted by flybynight View Post
    Hey guys:
    Had one problem with my math homework:

    \int \frac{y^2}{25+y^2} dy

    So I tried y=5\tan x

    So then: dy=5\sec^2xdx,

    So the integration appears as: \int \frac{25\tan^2x}{25(1+\tan^2x)}(5\sec^2x)dx

    Which simplifies to: \int 5\tan^2x dx, which is where I get stuck.

    Any advice on moving forward, or a different, easier method?

    Peter
    Make life easy for yourself by noting that \displaystyle \frac{y^2}{y^2 + 25} = \frac{(y^2 + 25) - 25}{y^2 + 25} = 1 - \frac{1}{y^2 + 25}.

    On the other hand, if you're determined to know how to find \displaystyle \int \tan^2 (x) \, dx then you should note that \tan^2 (x) = \sec^2 (x) - 1 (at which point it helps to know what the derivative of \tan (x) is).
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  3. #3
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    Here's what I did:
    \int 1-\frac{25}{y^2+25} dy

    y=5 \tan(x)

    dy=5 \sec^2(x) dx

    \int 1-\frac{25}{25(\tan^2(x))} 5 \sec^2(x) dx

    \int 1-\frac{5\sec^2(x)}{\sec^2(x)} dx

    \int 1-5 dx

    -4y=-4\arctan(\frac{y}{5})

    However, the answer in the textbook (confirmed by Mathematica) is y-5\arctan(\frac{y}{5})

    Any help?

    Peter
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  4. #4
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    Your algebra is a bit messy.

    First, use linearity to pull out the 1 term from the integral and get

    \displaystyle I = \int \left( 1 - \frac{25}{y^2 + 25} \right) dy = \int 1dy - 25 \int \frac{dy}{y^2 + 25} = \int 1dy - \frac{25}{25} \int  \left( \frac{1}{tan^2x + 1} \right) \cdot 5sec^2xdx

    Remembering that tan^2x + 1 = sec^2x and thus \frac{1}{tan^2x + 1} = cos^2x you get

    \displaystyle  I = y - \int 5dx = y -  5x = y - 5tan^{-1} \left(\frac{y}{5} \right)

    Your mistake was that from the first integral to the second (after performing the substitution), you multiplied only the \frac{25}{y^2+25} factor by 5sec^2xdx instead of the whole integrand. You also forgot a +25 in the denominator there.
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  5. #5
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    Oh I see. I can't believe that I over looked that. Thank you very much!
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