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Math Help - Super challenging definite Integral

  1. #1
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    Super challenging definite Integral

    I need to solve:

    \int_{-\pi \over 2}^{\pi \over 2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx

    This integral is beyond my capabilities! This was given to me as challenge by my classmate. According to him the answer is \pi \over 4.

    What should I do to solve it?
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  2. #2
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    Lightbulb Re: Super challenging definite Integral

    Quote Originally Posted by Ozymandias View Post
    I need to solve:

    \int_{-\pi \over 2}^{\pi \over 2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx

    This integral is beyond my capabilities! This was given to me as challenge by my classmate. According to him the answer is \pi \over 4.

    What should I do to solve it?
    This is a very tricky question. It is not as difficult as it appears! You will have to recall the properties of definite integrals.

    Let I= \int_{-\pi \over 2}^{\pi \over 2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx

    1. Use the identity : \int_{-a}^{a} f(x) dx= \int_{0}^{a}[f(x)+f(-x)]dx

    I=\int_{0}^{\pi \over 2} \left( \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )} + \frac{1}{2007^{-x}+1}\cdot \frac{\sin^{2008}(-x)}{\sin^{2008}(-x)+\cos^{2008}(-x)}\right)dx

    \implies I= \int_{0}^{\pi \over 2}  \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx \quad [1]

    2. Use the identity \int_{0}^{a}f(x) dx = \int_{0}^{a}f(a-x) dx:

    I=\int_{0}^{\pi \over 2}  \frac{\sin^{2008}(\frac{\pi}{2}-x)}{\sin^{2008}(\frac{\pi}{2}-x)+\cos^{2008}(\frac{\pi}{2}-x)}dx

    \implies I= \int_{0}^{\pi \over 2}  \frac{\cos^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx \quad [2]

    3. Add [1] and [2]:

    2I= \int_{0}^{\pi \over 2}\frac{\cos^{2008}(x)+\sin^{2008}(x)}{\sin^{2008}  (x)+\cos^{2008}(x)}dx

    \implies 2I = \int_{0}^{\pi \over 2} dx

    \implies 2I = \frac{\pi}{2}

    \implies I = \frac{\pi}{4}
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  3. #3
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    Re: Super challenging definite Integral

    Quote Originally Posted by sbhatnagar View Post
    This is a very tricky question. It is not as difficult as it appears! You will have to recall the properties of definite integrals.

    Let I= \int_{-\pi \over 2}^{\pi \over 2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx

    1. Use the identity : \int_{-a}^{a} f(x) dx= \int_{0}^{a}[f(x)+f(-x)]dx

    I=\int_{0}^{\pi \over 2} \left( \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )} + \frac{1}{2007^{-x}+1}\cdot \frac{\sin^{2008}(-x)}{\sin^{2008}(-x)+\cos^{2008}(-x)}\right)dx

    \implies I= \int_{0}^{\pi \over 2}  \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx \quad [1]

    2. Use the identity \int_{0}^{a}f(x) dx = \int_{0}^{a}f(a-x) dx:

    I=\int_{0}^{\pi \over 2}  \frac{\sin^{2008}(\frac{\pi}{2}-x)}{\sin^{2008}(\frac{\pi}{2}-x)+\cos^{2008}(\frac{\pi}{2}-x)}dx

    \implies I= \int_{0}^{\pi \over 2}  \frac{\cos^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx \quad [2]

    3. Add [1] and [2]:

    2I= \int_{0}^{\pi \over 2}\frac{\cos^{2008}(x)+\sin^{2008}(x)}{\sin^{2008}  (x)+\cos^{2008}(x)}dx

    \implies 2I = \int_{0}^{\pi \over 2} dx

    \implies 2I = \frac{\pi}{2}

    \implies I = \frac{\pi}{4}
    You are not a mathematician Sbhatnagar, you are a magician! You made the entire integral just vanish into the air.
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  4. #4
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    Re: Super challenging definite Integral

    Quote Originally Posted by sbhatnagar View Post
    This is a very tricky question. It is not as difficult as it appears! You will have to recall the properties of definite integrals.

    Let I= \int_{-\pi \over 2}^{\pi \over 2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx

    1. Use the identity : \int_{-a}^{a} f(x) dx= \int_{0}^{a}[f(x)+f(-x)]dx

    I=\int_{0}^{\pi \over 2} \left( \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )} + \frac{1}{2007^{-x}+1}\cdot \frac{\sin^{2008}(-x)}{\sin^{2008}(-x)+\cos^{2008}(-x)}\right)dx

    \implies I= \int_{0}^{\pi \over 2}  \frac{\sin^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx \quad [1]

    2. Use the identity \int_{0}^{a}f(x) dx = \int_{0}^{a}f(a-x) dx:

    I=\int_{0}^{\pi \over 2}  \frac{\sin^{2008}(\frac{\pi}{2}-x)}{\sin^{2008}(\frac{\pi}{2}-x)+\cos^{2008}(\frac{\pi}{2}-x)}dx

    \implies I= \int_{0}^{\pi \over 2}  \frac{\cos^{2008}(x)}{\sin^{2008}(x)+\cos^{2008}(x  )}dx \quad [2]

    3. Add [1] and [2]:

    2I= \int_{0}^{\pi \over 2}\frac{\cos^{2008}(x)+\sin^{2008}(x)}{\sin^{2008}  (x)+\cos^{2008}(x)}dx

    \implies 2I = \int_{0}^{\pi \over 2} dx

    \implies 2I = \frac{\pi}{2}

    \implies I = \frac{\pi}{4}
    Thank you sooooo much!!
    I thought about using some properties of definite integrals but did not get that far...

    Once again, thank you!
    Happy New year!
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