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Math Help - Trigonometric Integral

  1. #1
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    Trigonometric Integral

    Hi. I'm working through some review questions in prep for the mid-term. This one has me scratching my head a little bit:

    \int{\frac{1}{1-sin(t)}}\,dt

    I would appreciate any hints. I'm not sure how to apply here the normal approach of using identies and substitution. I played around with a few techniques like adding one, and IBP, but this didn't seem to help.
    Last edited by infraRed; October 31st 2013 at 01:29 PM. Reason: spelling
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  2. #2
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    Re: Trigonometric Integral

    Hey infraRed.

    One that comes to mind hint-wise is to multiply the inside by (1+sin(t))/(1+sin(t)) which will make the denominator 1 - sin^2(t) = cos^2(t) and this will simplify to sec^2(t) - tan^2(t) which is integrable using standard results.
    Thanks from topsquark and infraRed
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  3. #3
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    Re: Trigonometric Integral

    Hello, infraRed!

    \int \frac{1}{1-\sin t}\,dt

    Multiply by \tfrac{1+\sin t}{1+\sin t}\!:\;\; \frac{1}{1-\sin t}\cdot\frac{1+\sin t}{1+\sin t} \;=\;\frac{1+\sin t}{1-\sin^2\!t} \;=\; \frac{1+\sin t}{\cos^2\!t}

    . . . . . . . . . . . =\;\frac{1}{\cos^2\!t} + \frac{\sin t}{\cos^2\!t} \;=\;\sec^2\!t + \sec t\tan t


    And you can integrate: . \int(\sec^2\!t + \sec t\tan t)\,dt

    Right?
    Thanks from infraRed
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    Re: Trigonometric Integral

    Quote Originally Posted by Soroban View Post
    Hello, infraRed!


    Multiply by \tfrac{1+\sin t}{1+\sin t}\!:\;\; \frac{1}{1-\sin t}\cdot\frac{1+\sin t}{1+\sin t} \;=\;\frac{1+\sin t}{1-\sin^2\!t} \;=\; \frac{1+\sin t}{\cos^2\!t}

    . . . . . . . . . . . =\;\frac{1}{\cos^2\!t} + \frac{\sin t}{\cos^2\!t} \;=\;\sec^2\!t + \sec t\tan t


    And you can integrate: . \int(\sec^2\!t + \sec t\tan t)\,dt

    Right?
    It might be easier for the OP to write \displaystyle \begin{align*} \int{\sec^2{(t)}\,dt} - \int{ \frac{-\sin{(t)}}{\cos^2{(t)}}\,dt} \end{align*} as there is a more obvious substitution.
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    Re: Trigonometric Integral

    Quote Originally Posted by Soroban View Post
    Hello, infraRed!


    Multiply by \tfrac{1+\sin t}{1+\sin t}\!:\;\; \frac{1}{1-\sin t}\cdot\frac{1+\sin t}{1+\sin t} \;=\;\frac{1+\sin t}{1-\sin^2\!t} \;=\; \frac{1+\sin t}{\cos^2\!t}

    . . . . . . . . . . . =\;\frac{1}{\cos^2\!t} + \frac{\sin t}{\cos^2\!t} \;=\;\sec^2\!t + \sec t\tan t


    And you can integrate: . \int(\sec^2\!t + \sec t\tan t)\,dt

    Right?
    Thanks all! Forgot about the multiply-by-1 technique.
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  6. #6
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    Re: Trigonometric Integral

    An alternative method for dealing with integrals involving trig functions like this is to use the half-angle substitution

    u = \tan(t/2).

    That gets you \tan(t)=\frac{2u}{1-u^{2}}, \quad \sin(t)=\frac{2u}{1+u^{2}}, \quad \cos(t) = \frac{1-u^{2}}{1+u^{2}}, and dt = \frac{2du}{1+u^{2}}.

    For this particular example it's not as neat as the routine already demonstrated, but there are times when it's the thing to do.
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  7. #7
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    Re: Trigonometric Integral

    Hint. Final answer should come to 2sin(t/2)/(cos(t/2)-sin(t/2)) + C
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    Re: Trigonometric Integral

    Hint. Final answer should come to 2sin(t/2)/(cos(t/2)-sin(t/2)) + C
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