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Math Help - Integral of Trigonometric Functions

  1. #1
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    Integral of Trigonometric Functions

    Hi, sorry for not fixing my profile but I'll get right on it after posting this problem. I'm sure there is a rule about Latex here but forgive me if I will post the problem through an image. I hope that is okay.

    Here is my problem:



    I do not know where to start or perhaps I could not wrap around my head in this problem. I tried thinking of ways but I really can't and I am stumped. Our current and past lessons were Inverse Trigonometric Functions and Powers of Sine, Cosine, Tangents (Integrals, of course). I even tried Integration by parts but I still can't start how. So, this is my last resort.

    Please help
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  2. #2
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    \frac{\cot{\theta}}{\csc{\theta} + 4\sin{\theta}} = \frac{\frac{\cos{\theta}}{\sin{\theta}}}{\frac{1}{  \sin{\theta}} + 4\sin{\theta}}

     = \frac{\cos{\theta}}{1 + 4\sin^2{\theta}}


    So \int{\frac{\cot{x}}{\csc{x} + 4\sin{x}}\,dx} = \int{\frac{\cos{x}}{1 + 4\sin^2{x}}\,dx}.

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

    \int{\frac{1}{1 + 4u^2}\,du}.


    Now we need to make a trigonometric substitution u = \frac{1}{2}\tan{\theta} so that du = \frac{1}{2}\sec^2{\theta}\,d\theta and \theta = \arctan{2u} and the integral becomes

    \int{\frac{1}{1 + 4\left(\frac{1}{2}\tan{\theta}\right)^2}\left(\fra  c{1}{2}\sec^2{\theta}\right)\,d\theta}

     = \frac{1}{2}\int{\frac{\sec^2{\theta}}{1 + \tan^2{\theta}}\,d\theta}

     = \frac{1}{2}\int{\frac{\sec^2{\theta}}{\sec^2{\thet  a}}\,d\theta}

     = \frac{1}{2}\int{1\,d\theta}

     = \frac{1}{2}\theta + C

     = \frac{1}{2}\arctan{2u} + C

     = \frac{1}{2}\arctan{(2\sin{x})} + C.
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  3. #3
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    Follow up question,

    Where did you get u = \frac{1}{2}\tan{\theta} from if u = \sin{x}?
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  4. #4
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    It's just a renaming...
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  5. #5
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    I see, I'm just not accustomed to that kind of substitution. Trigonometric substitution was taught in a different way in what I saw in your post.

    If a^2+u^2 in substitution, shouldn't it be u = (a)tan{\theta}

    or you can substitute anything?

    I got the first part by the way and thanks for that, the second part is a little bit hazy and I really want to learn more.
    Last edited by gamedok; August 15th 2010 at 04:52 AM. Reason: Added something.
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  6. #6
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    If you wanted to do it that way, you would have to take out a factor of \frac{1}{4} to get it of that form.

    Of course then, you would have an extra factor to take out after you've made the substitution. The way I posted avoids taking out 2 extra common factors...
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  7. #7
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    If I have followed my formula, I would've ended with this:

    \int{\frac{\sec^2{\theta}}{1 + (4)\tan^2{\theta}}\,d\theta}

    I really appreciate your shortcut but I really need the longer method if that's alright. I really want to understand this better in order for me to teach it to my classmates as well.

    Where do I go from here?
    \int{\frac{\sec^2{\theta}}{1 + (4)\tan^2{\theta}}\,d\theta}
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  8. #8
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    No, that's not correct. Like I said, if you want to use the longer method you need to take out the common factor of \frac{1}{4} in the denominator, because you need it to be of the form a^2+ u^2, not a^2 + 4u^2...
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  9. #9
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    I'm going to sound a total newbie but... is this correct?

    EQUATION ERASED. WRONG EQUATION.

    ---------------------
    Oh crap, I just forgot factoring. =_=

    Still sorting it out.
    Last edited by gamedok; August 15th 2010 at 05:50 AM. Reason: Sorry just forgot something.
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  10. #10
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    1+4u^2

    Factoring this, I will have

    4 (\frac{1}{2}^2+u^2)

    So basically, that's where you got:
    u = \frac{1}{2}\tan{\theta}

    Am I right?
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  11. #11
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    Correct.

    And then after you have simplified this, you will need to factorise again to get 1 + \tan^2{\theta} which simplifies to \sec^2{\theta}.
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  12. #12
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    Thank you very much! Thanks for being patient with me.
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