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Math Help - integration of a cos function

  1. #1
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    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!!
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  2. #2
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    Is the t supposed to be there? Otherwise you have 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
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  3. #3
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    Hello, littlepig!

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


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

    Multiply by \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}

    . . =\;\;\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: . \int(\csc^2\!x - \csc x\cot x)\,dx \;\;=\;\;-\cot x + \csc x + C

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  4. #4
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    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 =(
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  5. #5
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    <br /> <br />
\int\frac{1}{1+t\cos x}\,dx<br />

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

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

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

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

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

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

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

    \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!
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  6. #6
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    Hello,

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

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

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

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

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

    So finally, 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 \pi/2

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

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

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

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

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


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


    Therefore :

    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}

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




    Note that everything is okay because |t|<1
    Last edited by Krizalid; May 29th 2009 at 01:23 PM.
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