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Another trig sub.

  1. Feb 9, 2005 #1
    i need to find the appropriate trigonometric substitution for this problem. i dont know why, but im only having problems finding the right sub. i can do the rest pretty easily. please help me with this:

    [tex]\int \frac{x^2}{sqrt(4x^2+8)}[/tex]

    here's what i done:

    sqrt(4(x^2+2)) ----> a*tan(theta)

    so i'm thinking that x should be equal to sqrt(2)/2*tan(theta), but it's incorrect. can someone help? (i know it's the wrong sub because i'm doing my hw online and it askes me for the trig sub first to check)
    Last edited: Feb 9, 2005
  2. jcsd
  3. Feb 9, 2005 #2
    Why 1/sqrt(2) and not just sqrt(2)?

  4. Feb 9, 2005 #3


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    Homework Helper

    How about a substitutiton involving [itex] \sinh [/itex] ...?After all,it's still a trigonometric function,except,that it's not circular.


    P.S.It should give the result immediately.
  5. Feb 9, 2005 #4
    can i try?

    \int_{}^{} {\frac{{x^2 }}{{\sqrt {4x^2 + 8} }}} dx \\ \\
    = \int_{}^{} {\frac{{x^2 }}{{2\sqrt {x^2 + 2} }}} dx \\
    [tex] x = \sqrt 2 \tan \theta [/tex]

    \int_{}^{} {\frac{{2\tan ^2 \theta \sqrt 2 \sec ^2 \theta }}{{2\sqrt 2 \sec \theta }}d\theta } \\
    = \int_{}^{} {\tan ^2 \theta } \sec \theta d\theta \\
    = \int_{}^{} {\sec \theta d \sec \theta } \\

  6. Feb 9, 2005 #5
    you know that

    [tex] 1 + tan^2\theta = sec^2 \theta [/tex]

    Use the subtitution, [tex] x = \sqrt{2} tan \theta [/tex]
  7. Feb 9, 2005 #6
    Oh I see when I finished typing the latex, somebody posted the solution.
  8. Feb 9, 2005 #7
    ah thanks for the help. dont know why i added the 1/2.

    we havent gone over sinh yet and dont think we ever will, so it'll be better if i dont use it.
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