Results 1 to 6 of 6

Thread: integration of a cos function

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    3

    integration of a cos function

    Hey, I was wondering if anyone could help me with the following integration:

    integrate 1/(1+tcosx) wrt x from 0 to pi. I know you have to make the substitution u=tan(x/2), but I've been trying for ages and can't seem to get the answer. =(

    Thanks in advance!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    Is the $\displaystyle t$ supposed to be there? Otherwise you have $\displaystyle u=\tan\frac{x}{2},\ du=\frac{dx}{2\cos^2 \frac{x}{2}}=\frac{dx}{1+\cos x} \implies dx=(1+\cos x)du$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, littlepig!

    Is that $\displaystyle "t"$ a typo? .If it is, the problem is quite simple.
    . . Otherwise, it's a nightmare!


    $\displaystyle \int\frac{1}{1+\cos x}\,dx$

    Multiply by $\displaystyle \frac{1-\cos x}{1-\cos x}\!:\quad \frac{1}{1 + \cos x}\cdot{\color{blue}\frac{1-\cos x}{1-\cos x}} \;\;=\;\;\frac{1-\cos x}{1-\cos^2\!x} \;\;=\;\;\frac{1-\cos x}{\sin^2\!x}$

    . . $\displaystyle =\;\;\frac{1}{\sin^2\!x} - \frac{\cos x}{\sin^2\!x} \;\;=\;\;\frac{1}{\sin^2\!x} - \frac{1}{\sin x}\!\cdot\!\frac{\cos x}{\sin x} \;\;=\;\;\csc^2\!x - \csc x\cot x$


    And we have: .$\displaystyle \int(\csc^2\!x - \csc x\cot x)\,dx \;\;=\;\;-\cot x + \csc x + C$

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2008
    Posts
    3
    The 't' isn't a typo. But it's given that |t|<1. The answer according to the notes is supposed to be pi/root(1-t^2), I just don't know how to get to it =(
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    $\displaystyle

    \int\frac{1}{1+t\cos x}\,dx
    $

    $\displaystyle \int \frac{1}{1+t \cos x}.\frac{1-t \cos x}{1-t \cos x}~dx$

    $\displaystyle \int \frac{1-t \cos x}{1-t^2 \cos^2 x}~dx$

    Let $\displaystyle u=1-t^2 \cos^2 x \Rightarrow \frac{du}{dx}=2t^2 \cos x \sin x $

    Notice that $\displaystyle u=1-t^2(1-\sin^2x)=1-t^2+t^2 \sin^2 x$

    $\displaystyle \frac{du}{dx}=2t^2 \left( \frac{1-u}{t^2} \right) \left( \frac{u+t^2-1}{t^2} \right)$

    $\displaystyle \int \frac{1-t \left( \frac{1-u}{t^2} \right)}{u} \frac{t^4 }{2t^2(1-u)(u+t^2-1)}~du$

    $\displaystyle \frac{1}{2t^2}\int \frac{t^4-t^3(1-u)}{u(1-u)(u+t^2-1)}~du$

    $\displaystyle \frac{t}{2} \int \frac{t-1+u}{u(1-u)(u+t^2-1)}~du$

    gah! this integral is harder than I first anticipated!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Let $\displaystyle I=\int_0^\pi \frac{1}{1+t\cos(x)} ~dx$

    By substituting $\displaystyle u=\pi-x$, we have :

    $\displaystyle I=-\int_\pi^0 \frac{1}{1-t\cos(u)} ~du$ (because $\displaystyle \cos(x)=\cos(\pi-u)=-\cos(u)$)

    So $\displaystyle I=\int_0^\pi \frac{1}{1-t\cos(x)} ~dx$ (u was just a dummy variable).

    Then
    $\displaystyle \begin{aligned}2I
    &=\int_0^\pi \frac{1}{1-t\cos(x)}+\frac{1}{1+t\cos(x)} ~dx \\
    &=\int_0^\pi \frac{1+t\cos(x)+1-t\cos(x)}{1-t^2\cos^2(x)} ~dx \\
    &=\int_0^\pi \frac{2}{1-t^2\cos^2(x)} ~dx \end{aligned}$

    So finally, $\displaystyle I=\int_0^\pi \frac{1}{1-t^2\cos^2(x)} ~dx$

    I have to think how to continue...



    I think I finally found it (after multiple careless mistakes )

    You can note that this latter integrand is symmetric with respect to $\displaystyle \pi/2$

    Hence $\displaystyle I=2\int_0^{\pi/2} \frac{1}{1-t^2\cos^2(x)} ~dx$

    We know that $\displaystyle \cos^2(x)=\frac{1+\cos(2x)}{2}$. So :

    $\displaystyle I=4\int_0^{\pi/2} \frac{1}{(2-t^2)-t^2\cos(2x)} ~dx$

    Now use Weierstrass substitution : $\displaystyle \cos(2x)=\frac{1-u^2}{1+u^2}$, where $\displaystyle u=\tan(x)$. --> $\displaystyle dx=\frac{du}{1+u^2}$

    $\displaystyle \begin{aligned}\Rightarrow I
    &=4\int_0^\infty \frac{du}{(2-t^2)(1+u^2)-t^2(1-u^2)} \\
    &=4\int_0^\infty \frac{du}{2-t^2+2u^2-u^2t^2-t^2+u^2t^2} \\
    &=4\int_0^\infty \frac{du}{2-2t^2+2u^2} \\
    &=2\int_0^\infty \frac{du}{(1-t^2)+u^2} \end{aligned}$


    But then, we know that an antiderivative of $\displaystyle \frac{1}{a+u^2}$ is $\displaystyle \frac{1}{\sqrt{a}} \cdot \arctan\left(\tfrac{u}{\sqrt{a}}\right)$


    Therefore :

    $\displaystyle I=\frac{2}{\sqrt{1-t^2}} \cdot \left.\arctan\left(\tfrac{u}{\sqrt{1-t^2}}\right)\right|_{u=0}^{u=\infty}=\frac{2}{\sqr t{1-t^2}} \cdot\frac{\pi}{2}$

    $\displaystyle \Longrightarrow \boxed{I=\frac{\pi}{\sqrt{1-t^2}}}$




    Note that everything is okay because $\displaystyle |t|<1$
    Last edited by Krizalid; May 29th 2009 at 01:23 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] integration of box function
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Jan 25th 2011, 03:36 PM
  2. Integration of a function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Aug 28th 2010, 01:23 AM
  3. Integration of function
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Dec 20th 2009, 03:39 AM
  4. Function of a Function Quotient Integration
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Jul 29th 2008, 10:13 AM
  5. Function of a function Quotient integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jul 29th 2008, 08:37 AM

Search Tags


/mathhelpforum @mathhelpforum