Results 1 to 12 of 12

Thread: Integral of Trigonometric Functions

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    9

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    $\displaystyle \frac{\cot{\theta}}{\csc{\theta} + 4\sin{\theta}} = \frac{\frac{\cos{\theta}}{\sin{\theta}}}{\frac{1}{ \sin{\theta}} + 4\sin{\theta}}$

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


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

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

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


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

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

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

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

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

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

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

    $\displaystyle = \frac{1}{2}\arctan{(2\sin{x})} + C$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2010
    Posts
    9
    Follow up question,

    Where did you get $\displaystyle u = \frac{1}{2}\tan{\theta}$ from if $\displaystyle u = \sin{x}$?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    It's just a renaming...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2010
    Posts
    9
    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 $\displaystyle a^2+u^2$ in substitution, shouldn't it be $\displaystyle u = (a)tan{\theta}$

    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; Aug 15th 2010 at 04:52 AM. Reason: Added something.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    If you wanted to do it that way, you would have to take out a factor of $\displaystyle \frac{1}{4}$ 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...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2010
    Posts
    9
    If I have followed my formula, I would've ended with this:

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

    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?
    $\displaystyle \int{\frac{\sec^2{\theta}}{1 + (4)\tan^2{\theta}}\,d\theta}$
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    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 $\displaystyle \frac{1}{4}$ in the denominator, because you need it to be of the form $\displaystyle a^2+ u^2$, not $\displaystyle a^2 + 4u^2$...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Aug 2010
    Posts
    9
    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; Aug 15th 2010 at 05:50 AM. Reason: Sorry just forgot something.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Aug 2010
    Posts
    9
    $\displaystyle 1+4u^2$

    Factoring this, I will have

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

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

    Am I right?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Correct.

    And then after you have simplified this, you will need to factorise again to get $\displaystyle 1 + \tan^2{\theta}$ which simplifies to $\displaystyle \sec^2{\theta}$.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Aug 2010
    Posts
    9
    Thank you very much! Thanks for being patient with me.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Aug 29th 2010, 05:23 PM
  2. Replies: 2
    Last Post: Feb 19th 2010, 10:37 AM
  3. Replies: 2
    Last Post: Jan 13th 2010, 07:31 PM
  4. An integral with trigonometric functions
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Jun 6th 2009, 02:52 PM
  5. Integral with trigonometric functions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 5th 2008, 07:21 PM

Search Tags


/mathhelpforum @mathhelpforum