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  1. #1
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    integral

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

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

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

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

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

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

    The rest is trivial.
    Last edited by mr fantastic; Sep 19th 2009 at 12:43 AM. Reason: Restored original reply
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  3. #3
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    Quote Originally Posted by luobo View Post
    ---------------------
    Let $\displaystyle A=\int^{\pi/2}_{-\pi/2} \frac{1}{2009^{x}+1} \cdot \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx$ (1)

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

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

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

    substitute $\displaystyle x \to \frac{\pi}{2} - x$ to get $\displaystyle A=\int_0^{\frac{\pi}{2}} \frac{\cos^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx.$ adding the two results together gives us $\displaystyle A=\frac{\pi}{4}.$
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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 $\displaystyle A=\int_0^{\frac{\pi}{2}} \frac{\sin^{2010}x}{\sin^{2010}x + \cos^{2010}x} \ dx.$

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

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

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

    Therefore,
    $\displaystyle A=\frac{\pi}{4}$
    Last edited by mr fantastic; Sep 19th 2009 at 12:44 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,
    $\displaystyle A=\int^{+\infty}_{0} \frac{t^{2010}}{(1+t^{2010})(1+t^2)} \ dt$ (5)

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

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

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