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  1. #1
    Super Member Random Variable's Avatar
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    integral

     \int^{\pi/2}_{-\pi/2} \frac{1}{2009^{x}+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx
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  2. #2
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    Quote Originally Posted by Random Variable View Post
     \int^{\pi/2}_{-\pi/2} \frac{1}{2009^{x}+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx
    ---------------------
    Let  A=\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{x}+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx (1)

    Let x=-t, then
     A=-\int^{-\pi/2}_{\pi/2} \frac{1}{2009^{-t}+1} \cdot \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt =\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{-t}+1} \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt

    Therefore,
     A=\int^{\pi/2}_{-\pi/2} \frac{2009^x}{2009^x+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx (2)

    (1)+(2),
     2A=\int^{\pi/2}_{-\pi/2} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx (3)

    Therefore,
     A=\int^{\pi/2}_{0} \frac{\tan^{2010}x}{\tan^{2010}x + 1} \ dx (4)

    Let t=\tan x, then
     A=\int^{+\infty}_{0} \frac{t^{2010}}{(t^{2010} + 1)(1+t^2)} \ dt (5)

    The rest is trivial.
    Last edited by mr fantastic; September 19th 2009 at 01:43 AM. Reason: Restored original reply
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  3. #3
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    Quote Originally Posted by luobo View Post
    ---------------------
    Let  A=\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{x}+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx (1)

    Let x=-t, then
     A=-\int^{-\pi/2}_{\pi/2} \frac{1}{2009^{-t}+1} \cdot \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt =\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{-t}+1} \frac{\sin^{2010}t}{\sin^{2010}t + \cos^{2010}t} \ dt

    Therefore,
     A=\int^{\pi/2}_{-\pi/2} \frac{2009^x}{2009^x+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx (2)

    (1)+(2),
     2A=\int^{\pi/2}_{-\pi/2} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx (3)
    from hee you should countinue like this: since the integrand in (3) is an even function, we have A=\int_0^{\frac{\pi}{2}} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx.

    substitute x \to \frac{\pi}{2} - x to get A=\int_0^{\frac{\pi}{2}} \frac{\cos^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx. adding the two results together gives us A=\frac{\pi}{4}.
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    from hee you should countinue like this: since the integrand in (3) is an even function, we have A=\int_0^{\frac{\pi}{2}} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx.

    substitute x \to \frac{\pi}{2} - x to get A=\int_0^{\frac{\pi}{2}} \frac{\cos^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx. adding the two results together gives us A=\frac{\pi}{4}.
    We can also continue like the following as well. Per Equation (5) in my previous post,
    A=\int^{+\infty}_{0} \frac{t^{2010}}{(1+t^{2010})(1+t^2)} \ dt (5)

    Let t=\frac{1}{x}, then
    A=\int^{+\infty}_{0} \frac{1}{(1+t^{2010})(1+t^2)} \ dt (6)

    (5)+(6),
    2A=\int^{+\infty}_{0} \frac{1}{1+t^2}=\frac{\pi}{2} (7)

    Therefore,
    A=\frac{\pi}{4}
    Last edited by mr fantastic; September 19th 2009 at 01:44 AM. Reason: Restored original reply
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  5. #5
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    Quote Originally Posted by luobo View Post
    We can also continue like the following as well. Per Equation (5) in my previous post,
    A=\int^{+\infty}_{0} \frac{t^{2010}}{(1+t^{2010})(1+t^2)} \ dt (5)

    Let t=\frac{1}{x}, then
    A=\int^{+\infty}_{0} \frac{1}{(1+t^{2010})(1+t^2)} \ dt (6)

    (5)+(6),
    2A=\int^{+\infty}_{0} \frac{1}{1+t^2}=\frac{\pi}{2} (7)

    Therefore,
    A=\frac{\pi}{4}
    that' right.
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