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    Hi, I need help with a textbook question:

    "If I_n denotes \displaystyle \int_0^{\frac{\pi}{2}} e^x \cos^n (x) \mathrm{d}x, show that:

    (n^2+1)I_n = n(n-1)I_{n-2} -1, and state the values of n for which the reduction formula is valid."

    I have tried using integration by parts such that:

    I_n = \Left[ e^x \cos^n(x) \Right]_0^{\frac{\pi}{2}} - n \displaystyle \int_0^{\frac{\pi}{2}} e^x \sin(x) \cos^{n-1} (x) \mathrm{d}x

    but then I'm not sure how to eliminate the \sin(x) from the integral.

    eg let J = \displaystyle \int_0^{\frac{\pi}{2}} e^x \sin(x) \cos^{n-1} (x) \mathrm{d}x

    if I split the integral J into e^x \cos^{n-1}(x) and \sin(x) for further integration by parts:

    J = \left[ -e^x\cos^n(x) \right]_0^{\frac{\pi}{2}} + (n-1)\displaystyle \int_0^{\frac{\pi}{2}} ( e^x \cos^{n-1}(x) \sin(x) + e^x \cos^n{x}) \mathrm{d}x

     \Rightarrow J = 1 + (n-1)J + (n-1)I_n

    \Rightarrow (n-2) J = -1 -(n-1)I_n

    And then I can't make much progress. Have I taken the wrong approach/made a mistake in my working, or is there some further insight to help carry on from here? Thanks
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    Try splitting into e^x and sin x cos^(n-1) x
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    (Original post by DFranklin)
    Try splitting into e^x and sin x cos^(n-1) x
    Thanks, it worked out in the end
 
 
 
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