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Math Help - Trig Integral

  1. #1
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    Trig Integral

    \int_{0}^{ \frac{\pi}{2}} \frac{1}{(1+ \sin x)(1+ \cos x)} dx

    My first attempt was to multiply the fraction by \frac{(1- \sin x)(1- \cos x)}{(1- \sin x)(1- \cos x)}, but I didn't get very far with it. Not too sure on how to involve the limits usefully.

    Bobak

    Edit: i posted the wrong limit
    Last edited by bobak; June 25th 2008 at 01:25 PM.
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  2. #2
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    \int_{0}^{\pi}\frac{1}{(1+sin(x))(1+cos(x))}dx

    You can try letting x=2tan^{-1}(u), , \;\ u=tan(x/2), \;\ du=\frac{1}{1+cos(x)}dx

    Make the subs and get:

    \int_{0}^{\infty}\frac{u^{2}+1}{u^{2}+2u+1}du

    This may look bad, but it can be expanded to:

    \int_{0}^{\infty}\left[\frac{2}{(u+1)^{2}}-\frac{2}{u+1}+1\right]du

    It is divergent.
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  3. #3
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    Hello

    Quote Originally Posted by bobak View Post
    \int_{0}^{ \pi} \frac{1}{(1+ \sin x)(1+ \cos x)} dx

    My first attempt was to multiply the fraction by \frac{(1- \sin x)(1- \cos x)}{(1- \sin x)(1- \cos x)}, but I didn't get very far with it. Not too sure on how to involve the limits usefully.

    Bobak
    Let me try...

    We know the formulae :

    \cos x=\frac{1-t^2}{1+t^2} and \sin x=\frac{2t}{1+t^2}, where t=\tan \frac x2


    Therefore 1+\cos x=\frac{2}{1+t^2} and 1+\sin x=\frac{(t+1)^2}{t^2+1}

    ---> (1+\cos x)(1+\sin x)=\frac{2(t+1)^2}{(t^2+1)^2}

    \frac{dt}{dx}=\frac 12 \cdot (1+\tan^2 \frac x2)=\frac 12 (1+t^2) \implies dx=\frac{2dt}{1+t^2}

    If x=0, \tan \frac x2=0.

    If x \to \pi^-, \tan \frac x2 \to +\infty

    ---> I=\int_0^\infty \frac{t^2+1}{(t+1)^2} dt

    I=\int_0^\infty 1-\frac{2t}{t^2+2t+1} dt

    I=\int_0^\infty 1-\frac{2t+2}{t^2+2t+1}+\frac{2}{(t+1)^2} dt


    I=\left[t-2\ln(t+1)-\frac{2}{t+1}\right]_0^\infty

    This is divergent...
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  4. #4
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    Quote Originally Posted by galactus View Post
    \int_{0}^{\pi}\frac{1}{(1+sin(x))(1+cos(x))}dx

    You can try letting x=2tan^{-1}(u), , \;\ u=tan(x/2), \;\ du=\frac{1}{1+cos(x)}dx

    Make the subs and get:

    \int_{0}^{\infty}\frac{u^{2}+1}{u^{2}+2u+1}du

    This may look bad, but it can be expanded to:

    \int_{0}^{\infty}\left[\frac{2}{(u+1)^{2}}-\frac{2}{u+1}+1\right]du

    It is divergent.
    I posted the wrong limits but I get it now. thanks.

    Bobak
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