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Math Help - Integration by trig substitution

  1. #1
    Newbie madmartigano's Avatar
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    Integration by trig substitution

    I need to show that \int {\frac{x}{{{x^2}+4}}dx} <br />
=\frac{1}{2}\ln \left( {{x^2} + 4} \right) + c<br />
using trig substitution.

    I've tried the following:

    \begin{array}{l}<br />
 x = 2\tan \theta ,   dx = 2{\sec ^2}\theta d\theta  \\ <br />
  \\ <br />
 \int {\frac{{4\tan \theta {{\sec }^2}\theta }}{{4{{\tan }^2}\theta  + 4}}d\theta  = \int {\frac{{\tan \theta {{\sec }^2}\theta }}{{{{\sec }^2}\theta }}d\theta  = \int {\tan \theta d\theta  = \ln \left| {\sec \theta } \right| + c} } }  \\ <br />
 \end{array}<br />

    \theta  = {\tan ^{ - 1}}\frac{x}{2},   \sec \theta  = \frac{{\sqrt {{x^2} + 4} }}{2},<br />

    but \int {\frac{x}{{{x^2} + 4}}dx \ne \ln \left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right)}  + c<br />

    Where am I going wrong?
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  2. #2
    Junior Member
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    Quote Originally Posted by madmartigano View Post
    I need to show that \int {\frac{x}{{{x^2}+4}}dx} <br />
=\frac{1}{2}\ln \left( {{x^2} + 4} \right) + c<br />
using trig substitution.

    I've tried the following:

    \begin{array}{l}<br />
 x = 2\tan \theta ,   dx = 2{\sec ^2}\theta d\theta  \\ <br />
  \\ <br />
 \int {\frac{{4\tan \theta {{\sec }^2}\theta }}{{4{{\tan }^2}\theta  + 4}}d\theta  = \int {\frac{{\tan \theta {{\sec }^2}\theta }}{{{{\sec }^2}\theta }}d\theta  = \int {\tan \theta d\theta  = \ln \left| {\sec \theta } \right| + c} } }  \\ <br />
 \end{array}<br />

    \theta  = {\tan ^{ - 1}}\frac{x}{2},   \sec \theta  = \frac{{\sqrt {{x^2} + 4} }}{2},<br />

    but \int {\frac{x}{{{x^2} + 4}}dx \ne \ln \left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right)}  + c<br />

    Where am I going wrong?
    That's a strange question - you don't need any substitution to show that, it's just kind of obvious.
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  3. #3
    Newbie madmartigano's Avatar
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    The problem is strange but necessary. I would prefer to use u=x^2+4 and be done with it, but unfortunately I can't. I've been working at this problem for an hour; I don't know where I'm erring.
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  4. #4
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by madmartigano View Post
    I need to show that \int {\frac{x}{{{x^2}+4}}dx} <br />
=\frac{1}{2}\ln \left( {{x^2} + 4} \right) + c<br />
using trig substitution.

    I've tried the following:

    \begin{array}{l}<br />
 x = 2\tan \theta ,   dx = 2{\sec ^2}\theta d\theta  \\ <br />
  \\ <br />
 \int {\frac{{4\tan \theta {{\sec }^2}\theta }}{{4{{\tan }^2}\theta  + 4}}d\theta  = \int {\frac{{\tan \theta {{\sec }^2}\theta }}{{{{\sec }^2}\theta }}d\theta  = \int {\tan \theta d\theta  = \ln \left| {\sec \theta } \right| + c} } }  \\ <br />
 \end{array}<br />

    \theta  = {\tan ^{ - 1}}\frac{x}{2},   \sec \theta  = \frac{{\sqrt {{x^2} + 4} }}{2},<br />

    but \int {\frac{x}{{{x^2} + 4}}dx \ne \ln \left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right)}  + c<br />

    Where am I going wrong?
    You're not going wrong. Use properties of logarithms to see that \ln \left( {\frac{{\sqrt {{x^2} + 4} }}{2}} \right) = \ln(x^2 + 4)^{1/2} - \ln2 = \tfrac12\ln(x^2+4) - \ln2, which differs from the given answer only by a constant of integration.

    I have to agree with JeWiSh that this is a really daft way to calculate this integral. Much better would be to make the substitution y=x^2+4, leading immediately to the answer \tfrac12\ln y + C.
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  5. #5
    Newbie madmartigano's Avatar
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    Ahhh. I separated the log the same as you, but I disregarded the answer because of the additional ln 2. Thank you for clarifying things.
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