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Math Help - Finding a definite integral

  1. #1
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    Finding a definite integral

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


    Thanks.
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  2. #2
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    Quote Originally Posted by maliky View Post
    The question is:
    Find the definite integral:
     \int\limits_0^{\pi/2} \frac{1}{1+cos(\alpha)cos(x)}dx
    ( \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:
    \frac{\alpha}{\sin(\alpha)}


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

    Click Show Steps
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  3. #3
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    Universal substitution is probably the way to go.

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


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

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

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


    so  \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

     = \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}}

     = \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}}

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

     = \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  \alpha = \frac{\pi}{2} , then we have  \arctan 1 = \frac{\pi}{4} = \frac{\alpha}{2}

    so  \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos \alpha \cos x} \ dx = \frac{2}{\sin \alpha} \frac{\alpha}{2} = \frac{\alpha}{\sin \alpha}
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