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Math Help - integrate this function :S

  1. #1
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    integrate this function :S

    Dont know where to start..

    Integrate

    < avoid this question >(a^px)*(b^px)

    also integrate this function

    (sinx + cosx)/(sin^4x + cos^2x)

    please dont send me a wolfram alpha link.... -_-
    can we do this question witout using the universal substitution tanx/2 = t ?
    Last edited by ice_syncer; November 14th 2010 at 01:18 PM. Reason: because I got the answer to the first one
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    Quote Originally Posted by ice_syncer View Post
    Dont know where to start..

    Integrate

    (a^px)*(b^px)
    a^{px} \cdot b^{px} = (ab)^{px}

    let c = ab ...

    \displaystyle \int c^{px} \, dx = \frac{c^{px}}{p \ln{c}} + C = \frac{(ab)^{px}}{p \ln(ab)}+C
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    sorry thats the wrong question

    actually it is integrate (a^px)*(b^qx)
    sorryy
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    Quote Originally Posted by ice_syncer View Post
    sorry thats the wrong question

    actually it is integrate (a^px)*(b^qx)
    sorryy
    a^{px} \cdot b^{qx} = (a^p b^q)^x

    since a, b, p, and q are all constants ...

    let c = a^p b^q

    \displaystyle \int c^x \, dx = \frac{c^x}{\ln{c}} + C<br />

    back substitute ...

    \displaystyle \frac{(a^p b^q)^x}{\ln(a^p b^q)} + C
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    oh yeah, ok. thanks : )
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    Quote Originally Posted by ice_syncer View Post
    Dont know where to start..
    Integrate (a^px)*(b^px)

    (i) If you mean \int (a^px)(b^px)dx, then:

     \displaystyle\int (a^px)(b^px)dx=(ab)^{p}\int x^2dx=\dfrac{(ab)^px^3}{3}+C

    (ii) If you mean \int (a^{px})(b^{px})dx, then:

    \displaystyle\int (a^{px})(b^{px})dx=\displaystyle\int (ab)^{px}dx=\dfrac{1}{p\log (ab)}(ab)^{px}+C

    Regards.

    Edited: Sorry, when I submitted my answer, I didn't see the previous posts. Besides, I read b^{px} instead of b^{qx} .
    Last edited by FernandoRevilla; November 14th 2010 at 01:32 PM.
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    thanks
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  8. #8
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    the next question is hard for me becauseof complicated partial fractions after substituting, actually you get surd like stuff
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    Quote Originally Posted by ice_syncer View Post
    the next question is hard for me becauseof complicated partial fractions after substituting, actually you get surd like stuff
    I would guess so ... Wolfram gives a really ugly antiderivative in terms of an inverse hyperbolic tangent and inverse tangent.
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  10. #10
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    someone help me with this
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