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Math Help - Integrating inverse of trig products

  1. #1
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    Question Integrating inverse of trig products

    Hi
    I'm kind of hunting out of my depth relative to my current knowledge so the answer to this might be really obvious - please bear with me ...

    I'm trying to find
    \int \frac{1}{sin\theta cos\theta }d\theta

    I've tried integrating the inverse of a function
    Inverse Function Integration -- from Wolfram MathWorld
    but I ended up with an integral involving
    sin\left (sin\theta cos\theta  \right )cos\left (sin\theta cos\theta  \right )
    which didn't look very hopeful. It may be because I haven't officially learnt integrating an inverse yet ...

    I then tried integration by parts (again something which I had never heard of before), and I changed the original integral to
    \int sec\theta csc\theta \, d\theta
    to do so, but then when it came to chosing which of sec or csc to integrate and which to differentiate, either way seemed to end up with the 'output' integral containing piles of ln, cot, csc, tan all in arrangements I couldn't hope to integrate.

    Is there a technique(s) for integrating this which I don't know about yet? Any hints?
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  2. #2
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    \displaystyle  \int\frac{1}{\sin{x}\cos{x}}\;{dx}  = \int\frac{\cos^2{x}+\sin^2{x}}{\sin{x}\cos{x}}\;{d  x} \displaystyle = \int \frac{\cos{x}}{\sin{x}}\;d{x}-\int \frac{-\sin{x}}{\cos{x}}\;{dx} = \ln\left(\sin{x}\right)-\ln\left(\cos{x}\right)+k.
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  3. #3
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    An alternative:

    \displaystyle \int{\frac{1}{\sin{x}\cos{x}}\,dx} = \int{\frac{\cos{x}}{\sin{x}\cos^2{x}}\,dx}

    \displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin^2{x})}\,dx}


    Now make the substitution \displaystyle u = \sin{x} so that \displaystyle du = \cos{x}\,dx and the integral becomes

    \displaystyle \int{\frac{1}{u(1 - u^2)}\,du}

    which can be solved using Partal Fractions.
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  4. #4
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    Ah ha! Thank you very much my friends
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  5. #5
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    Hello, chug1!

    Here's a sneaky approach . . .


    \displaystyle \int \frac{1}{\sin\theta\cos\theta}\,d\theta

    \text{Divide numerator and denominator by }\cos^2\!\theta:\;\;\dfrac{\dfrac{1}{\cos^2\!\thet  a}} {\dfrac{\sin\theta\cos\theta}{\cos^2\!\theta}} \;=\;\dfrac{\sec^2\!\theta}{\tan\theta}

    \displaystyle \text{The integral becomes: }\;\int \frac{\sec^2\!\theta}{\tan\theta}\,d\theta \;=\;\ln|\tan\theta| + C

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  6. #6
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    Quote Originally Posted by Soroban View Post
    Here's a sneaky approach . . .
    Nice! I guess it's one of those integrals that can be evaluated in finitely many ways!

    4. Letting x = \arctan{t} gives:

    \begin{aligned}\displaystyle  \int\frac{1}{\sin{x}\cos{x}}\;{dx} & = \int\frac{2}{\sin{2x}}\;{dx}  = \int\frac{2}{(1+t^2)\times \frac{2t}{1+t^2}}\;{dt} \\&= \int\frac{1}{t}\;{dt} = \ln{t}+k = \ln(\tan{x})+k.\end{aligned}

    5. Letting t = \frac{\sin{x}}{\cos{x}} gives:

    \begin{aligned}\displaystyle  \int\frac{1}{\sin{x}\cos{x}}\;{dx} & = \int\frac{\cos^2{x}}{\sin{x}\cos{x}}\;{dt}  = \int\frac{\cos{x}}{\sin{x}}\;{dt} = \int\frac{1}{t}\;{dt} \\& = \ln{t}+k = \ln\left(\frac{\sin{x}}{\cos{x}}\right)+k  = \ln\left(\tan{x}\right)+k.\end{aligned}

    Notice that the above substitutions are actually equivalent!
    Last edited by TheCoffeeMachine; January 21st 2011 at 09:37 PM.
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  7. #7
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    An alternative :

    \displaystyle I =  \int\frac{1}{sin(x)cos(x)} \, dx = 2 \int \dfrac{1}{sin(2x)} \, dx = 2 \int \, csc(2x) \, dx

    u=2x \implies du=2dx \implies I = \int csc(u) \, du = ln|csc(u)-cot(u)| + C = ln|csc(2x)-cot(2x)| + C
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  8. #8
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    Just a side note - what the OP posted are not inverse trigonometric functions, they are reciprocal trigonometric functions...
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  9. #9
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    That's what I was about to say! (And will stick in my oar anyway!) It may be that the OP's problem was thinking that "inverse functions" ( sin^{-1}= arcsin, cos^{-1}= arccos) are the same as the reciprocals ( \frac{1}{sin}= csc, \frac{1}{cos}= sec) when he used "MathWorld" and so he entered the wrong functions. That's the problem with sites like that- they can't your thinking for you- GIGO!
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