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Math Help - simple integration

  1. #1
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    simple integration

    ok I know how integrals work and I have been using integrals to find area under the curves, I know integration by U sub and also by parts. but for some reason maybe it just slipped my mind but how do I integrate:
    integral tan (3x-1),x
    I started off by doing a u sub as u = (3x-1) and where i got 1/3du = dx,
    so my new equation became 1/3 int tan u du then i realized i don't know the integral of tan u. Can someone show me the steps using natural logs to the answer. I searched it on wolframaplpha but it gave me a log answer. Any help is appreciated.
    -Marcus
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: simple integration

    Quote Originally Posted by mjfuentes85 View Post
    ok I know how integrals work and I have been using integrals to find area under the curves, I know integration by U sub and also by parts. but for some reason maybe it just slipped my mind but how do I integrate:
    integral tan (3x-1),x
    I started off by doing a u sub as u = (3x-1) and where i got 1/3du = dx,
    so my new equation became 1/3 int tan u du then i realized i don't know the integral of tan u. Can someone show me the steps using natural logs to the answer. I searched it on wolframaplpha but it gave me a log answer. Any help is appreciated.
    -Marcus

    \int \tan(3x-1)dx=-\int \tan(1-3x)dx \overset{1-3x=t}{=}\frac{1}{3}\int \tan(t)dt=-\frac{1}{3} \log|\cos(t)|+C=\log|\cos(1-3x)|+C
    Last edited by Also sprach Zarathustra; June 13th 2011 at 10:23 PM.
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  3. #3
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    Re: simple integration

    The answer I was trying to get was ln |sec x | + c, which is the integral of tanx dx , could you show me how to get this answer ??
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: simple integration

    Quote Originally Posted by mjfuentes85 View Post
    The answer I was trying to get was ln |sec x | + c, which is the integral of tanx dx , could you show me how to get this answer ??

    \log\cos(1-3x)=\log\frac{1}{\frac{1}{cos(1-3x)}}=(\log\sec(1-3x))^{-1}=-\log\sec(1-3x)
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  5. #5
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    Re: simple integration

    Quote Originally Posted by Also sprach Zarathustra View Post
    \log\cos(1-3x)=\log\frac{1}{\frac{1}{cos(1-3x)}}=(\log\sec(1-3x))^{-1}=-\log\sec(1-3x)
    You really need to get into the habit of writing modulus signs around the stuff inside logarithms you get through integrating...
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    Re: simple integration

    Quote Originally Posted by mjfuentes85 View Post
    The answer I was trying to get was ln |sec x | + c, which is the integral of tanx dx , could you show me how to get this answer ??
    tan(x)= \frac{sin(x)}{cos(x)} so \int tan(x)dx= \int\frac{sin(x)}{cos(x)}dx. Let u= cos(x) so that du= -sin(x) and the integral becomes
    -\int \frac{du}{u}= -ln|u|+ C= -ln|cos(x)|+ C= ln|\frac{1}{cos(x)}|+ C= ln|sec(x)|+ C
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