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Math Help - Integrate problem

  1. #1
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    Integrate problem

    im sorry but i do not know how to use many of the math features to set up symbols and what not. so bare with me a little.

    integrate: tanX ln(cosX) dx

    so far i have:

    sinX
    ----- ln(cosX)dx
    cosX


    let u= cosX

    sinX
    ----- ln(u)dx
    u

    then:
    du = sinX ln(u) dx
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  2. #2
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    What about substituting u=ln(cos x) ?!!
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by reino17 View Post
    im sorry but i do not know how to use many of the math features to set up symbols and what not. so bare with me a little.

    integrate: tanX ln(cosX) dx

    so far i have:

    sinX
    ----- ln(cosX)dx
    cosX


    let u= cosX

    sinX
    ----- ln(u)dx
    u

    then:
    du = sinX ln(u) dx
    You dont really need to break that tan into sin and cos..

    \displaystyle \int \tan(x)\log(\cos(x)) \;dx

    Follow what Liverpool has said in the above post..

    let u=\log(\cos(x)) \rightarrow du\;=\;\dfrac{1}{\cos(x)} ({-\sin(x)}) dx = -\tan(x)\;dx \rightarrow dx = \dfrac{du}{\tan(x)}


    \displaystyle \int \tan(x)\log(\cos(x)) \;dx = \int\tan(x)\;u\;du\; \dfrac{-1}{\tan(x)} =-\int u\;du
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