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Math Help - Today's calculation of integral #11

  1. #1
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    Today's calculation of integral #11

    Let \alpha>0,\,\beta<1 and find \int_{0}^{\pi/2}{2\sec x\cdot \ln \left[ \frac{1+\beta \cos x}{1+\alpha \cos x} \right]\,dx}.

    (We can kill this little problem with a simple trick.)
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  2. #2
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    Hint (not actually a Hint): find an integral parameter.
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  3. #3
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    I have a solution.

    I don't know what an 'integral parameter' is, and the concept of differentiating inside the integral sign is fairly new to me, so bear with me.

    Let I_{n} = \int^{\frac{\pi}{2}}_{0} 2\sec x \ln(1 + n\cos x) \; \mathrm{d}x

    Then \frac{\mathrm{d}I_{n}}{\mathrm{d}{n}} = \int^{\frac{\pi}{2}}_{0} \frac{2}{1 + n\cos x} \; \mathrm{d}x

    Using the substitution t = \tan \frac{x}{2}, this is equal to 4\left[\frac{1}{\sqrt{1-n^{2}}} \tan^{-1} \left( t\sqrt{\frac{1-n}{1+n}}\right) \right]^{1}_{0}

    \Rightarrow I_{n} = \int  \frac{4}{\sqrt{1-n^{2}}}  \tan^{-1} \left(\sqrt{\frac{1-n}{1+n}}\right) \; \mathrm{d}n

    Noting that \frac{\mathrm{d}}{\mathrm{d}n} \tan^{-1} \left(\sqrt{\frac{1-n}{1+n}}\right) = - \frac{1}{2\sqrt{1-n^{2}}}, we have I_{n} = c - 4\left(\tan^{-1}\sqrt{\frac{1-n}{1+n}}\right)^{2}

    \Rightarrow I_{\beta} - I_{\alpha} = 4\left[\left(\tan^{-1}\sqrt{\frac{1-\alpha}{1+\alpha}}\right)^{2} - \left(\tan^{-1}\sqrt{\frac{1-\beta}{1+\beta}}\right)^{2} \right]

    This is equal to the integral that we have been asked to find.
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  4. #4
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    Okay, just express your answer in terms of arccos.

    When I get back at home I'll post another solution.
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  5. #5
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    \tan^{-1}\sqrt{\frac{1-x}{1+x}} = \cos^{-1}\sqrt{\frac{1+x}{2}} = \frac{\cos^{-1}x}{2}

    So the answer can be expressed as

    \left(\cos^{-1}\alpha\right)^{2} - \left(\cos^{-1}\beta\right)^{2}

    Which is much nicer.
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  6. #6
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    2\sec x\cdot \ln \left[ \frac{1+\beta \cos x}{1+\alpha \cos x} \right]

    2\sec x\cdot [\ln (1+\beta \cos x) - \ln (1+\alpha \cos x) ]

    \int^{\beta}_{\alpha} \frac{2}{1+y\cos x}~dy

    So the problem becomes,

    \int^{\frac{\pi}{2}}_{0}\int^{\beta}_{\alpha} \frac{2}{1+y\cos x}~dy~dx

    Now change the integration order and so on..
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  7. #7
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    Yes, that was the trick.
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