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Math Help - Reduction Formula

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    Reduction Formula

    Please help with this question. Thanks in advance.

    Q:
    I_n = \displaystyle\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x

    Prove that I_n = \frac{\pi ^n}{2^n} - n (n - 1) I_{n-2}, \ n \geq 2.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Air View Post
    Please help with this question. Thanks in advance.

    Q:
    I_n = \displaystyle\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x

    Prove that I_n = \frac{\pi ^n}{2^n} - n (n - 1) I_{n-2}, \ n \geq 2.
    Use parts twice
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Use parts twice
    Which should I take as u and \frac{\mathrm{d}v}{\mathrm{d}x} to be?
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Air View Post
    Which should I take as u and \frac{\mathrm{d}v}{\mathrm{d}x} to be?
    I would say cos(x) for u since \frac{D[\cos(x)]^2}{dy^2}=-\cos(x) which seems to be a part of the second term in the proof
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Air View Post
    Which should I take as u and \frac{\mathrm{d}v}{\mathrm{d}x} to be?
    u = x^n and du/dx = cos(x) seems to be the right choice. (the n(n - 1) part of the formula hints at this)
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    generally i will go for u = x^n and dv/dt = cos x. My reasoning is that if x^n is taken as dv/dt then the power of x will increase which brings you nowhere closer in solving the integral. However if you set u = x^n for each part the power of x decreases till x^0 (this gives an integration of a trigonometry which is simple to solve).
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  7. #7
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    Thanks all.

    I've done it and obtained the correct answer however my method was pretty long and slightly doubtful. Could someone also do this question so I can see if they do this 'Reduction Formula' method. I will be grateful if someone could help. Thanks in advance.
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  8. #8
    Moo
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    Hello,

    I don't think you could avoid it...

    I_n =\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x

    I'll name \frac{\pi}{2} y because it's really awful to write

    =[-x^n \sin x]_0^y + n \int_0^y x^{n-1} \sin x dx

    =y^n+n \underbrace{\left( [x^{n-1} \cos x]_0^y \right)}_{=0} + n(n-1) \underbrace{\int_0^y x^{n-2} \cos x dx}_{I_{n-2}}

    =\frac{\pi^n}{2^n}+n(n-1)I_{n-2}




    Is it your reduction formula or are you looking for the induction thing ?
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    Quote Originally Posted by Moo View Post
    Hello,

    I don't think you could avoid it...

    I_n =\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x

    I'll name \frac{\pi}{2} y because it's really awful to write

    =[-x^n \sin x]_0^y + n \int_0^y x^{n-1} \sin x dx

    =y^n+n \underbrace{\left( [x^{n-1} \cos x]_0^y \right)}_{=0} + n(n-1) \underbrace{\int_0^y x^{n-2} \cos x dx}_{I_{n-2}}

    =\frac{\pi^n}{2^n}+n(n-1)I_{n-2}




    Is it your reduction formula or are you looking for the induction thing ?
    No, it's the method you've done. I tend to write each stage so the method appears long but how you have done it has helped clarify the method. Thanks.
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  10. #10
    Moo
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    I know I've missed some steps, but it should add one or two more lines for each line..
    I hope this will really help
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