Results 1 to 3 of 3

Thread: Finding a definite integral

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    3

    Finding a definite integral

    The question is:
    Find the definite integral:
    $\displaystyle \int\limits_0^{\pi/2} \frac{1}{1+cos(\alpha)cos(x)}dx$
    ($\displaystyle \alpha$ is a constant)

    I've tried using the t-substitution, but I got quite a messy answer.
    Possibly there's some simplification involving the fact that the constant is written in terms of cos that I'm not seeing?

    The solution given is
    Spoiler:
    $\displaystyle \frac{\alpha}{\sin(\alpha)}$


    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by maliky View Post
    The question is:
    Find the definite integral:
    $\displaystyle \int\limits_0^{\pi/2} \frac{1}{1+cos(\alpha)cos(x)}dx$
    ($\displaystyle \alpha$ is a constant)

    I've tried using the t-substitution, but I got quite a messy answer.
    Possibly there's some simplification involving the fact that the constant is written in terms of cos that I'm not seeing?

    The solution given is
    Spoiler:
    $\displaystyle \frac{\alpha}{\sin(\alpha)}$


    Thanks.
    http://www.wolframalpha.com/input/?i...Cos[x]%29]

    Click Show Steps
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Universal substitution is probably the way to go.

    $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos \alpha \cos x} \ dx $


    let $\displaystyle t = \tan \Big(\frac{x}{2} \Big)$

    then $\displaystyle dx = \frac{2 \ dt}{1+t^{2}} $

    and $\displaystyle \cos x = \frac{1-t^{2}}{1+t^{2}} $


    so $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos \alpha \cos x} \ dx = \int_{0}^{1} \frac{\frac{2}{1+t^{2}}}{1+ \cos \alpha \frac{1-t^{2}}{1+t^{2}}} \ dt $

    $\displaystyle = \int^{1}_{0} \frac{2 \ dt}{1+\cos \alpha + t^{2} (1- \cos \alpha)} \ dt = \frac{2}{1+ \cos \alpha} \int^{1}_{0} \frac{dt}{1+ \Big(t \sqrt{\frac{1-\cos \alpha}{1+ \cos \alpha}}\Big)^{2}}$

    $\displaystyle = \frac{2 \sqrt{1+\cos \alpha}}{(1+\cos \alpha)\sqrt{1-\cos \alpha}} \int_{0}^{\sqrt{\frac{1-\cos \alpha}{1+ \cos \alpha}}} \frac{du}{1+u^{2}} $

    $\displaystyle = \frac{2}{\sqrt{1+\cos \alpha}\sqrt{1-\cos \alpha}} \arctan \Big( \sqrt{\frac{1-\cos \alpha}{1+ \cos \alpha}}\Big) $

    $\displaystyle = \frac{2}{\sqrt{1-\cos^{2}\alpha}} \arctan \Big( \sqrt{\frac{1-\cos \alpha}{1+ \cos \alpha}}\Big) = \frac{2}{\sin \alpha } \arctan \Big( \sqrt{\frac{1-\cos \alpha}{1+ \cos \alpha}}\Big) $


    now if $\displaystyle \alpha = \frac{\pi}{2} $, then we have $\displaystyle \arctan 1 = \frac{\pi}{4} = \frac{\alpha}{2} $

    so $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos \alpha \cos x} \ dx = \frac{2}{\sin \alpha} \frac{\alpha}{2} = \frac{\alpha}{\sin \alpha} $
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: May 29th 2011, 07:50 AM
  2. Replies: 4
    Last Post: Apr 13th 2011, 02:08 AM
  3. definite integral....finding the function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 25th 2009, 01:58 PM
  4. finding f for a definite integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Sep 20th 2009, 11:04 AM
  5. finding the derivative of a definite integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 3rd 2008, 02:34 PM

Search Tags


/mathhelpforum @mathhelpforum