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  1. #1
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    integration question help

    ok. so the book says to prove the reduction formula for:
    integral of cos^n x dx
    the says this equals 1/n cos^(n-1) x * sin x + ((n-1)/n) * integral of cos^n-2 x dx

    i tried to do this by breaking cos^n x into cos^n-1 * cos x
    i let u equal cos^n-1 x and dv = cos x dx
    for du i got -(n-1)(cos^n-2 x)(sin x) dx
    and for v i got sin x
    so the answer i got was cos^n-1 * sin x + (n-1) * the integral of cos^n-2 x * sin x dx

    what did i do wrong
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pcgamer03 View Post
    ok. so the book says to prove the reduction formula for:
    integral of cos^n x dx
    the says this equals 1/n cos^(n-1) x * sin x + ((n-1)/n) * integral of cos^n-2 x dx

    i tried to do this by breaking cos^n x into cos^n-1 * cos x
    i let u equal cos^n-1 x and dv = cos x dx
    for du i got -(n-1)(cos^n-2 x)(sin x) dx
    and for v i got sin x
    so the answer i got was cos^n-1 * sin x + (n-1) * the integral of cos^n-2 x * sin x dx

    what did i do wrong
    you started out well. see here and match up with your work to see where you went wrong
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  3. #3
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    Quote Originally Posted by pcgamer03 View Post
    ok. so the book says to prove the reduction formula for:
    integral of cos^n x dx
    the says this equals 1/n cos^(n-1) x * sin x + ((n-1)/n) * integral of cos^n-2 x dx
    Let I_n = \int \cos^n x dx.

    Then if n\geq 2 we can write, \int \cos^{n-1} x \cos^1 x dx

    Let, s'=\cos x \mbox{ and }t=\cos^{n-1} x
    Then,  s = -\sin x \mbox{ and }t ' = -(n-1)\cos^{n-2} \sin x

    Apply integration by parts, -\sin x \cos^{n-1} x + (n-1)\int \sin^2 x \cos^{n-2} x dx

    Now, \int \sin^2 x \cos^{n-2} xdx = \int (1-\cos^2 x)\cos^{n-2} dx = I_{n-2} - I_n

    Thus, I_n = -\sin x \cos^{n-1} x + (n-1)I_{n-2} - (n-1)I_n

    Solve, I_n = -\frac{1}{n}\sin x \cos^{n-1} x + \frac{(n-1)}nI_{n-2}

    There should not be a negative. But I am too lazy to check where is the mistake.
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