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Integration by parts expression

  1. Apr 16, 2008 #1
    Use integration by parts to express:

    I (n) = ∫(sin)^n (x) dx in terms of I (n-2)

    Let u = sinn-1 x
    du = (n-1)sinn-2 x cos x dx
    v = - cos x
    dv = sinx dx

    so integration by parts give:

    ∫〖sin〗^n x dx= -cos⁡〖x 〖sin〗^(n-1) x+(n-1) ∫〖sin〗^(n-2) 〗 x cos^2⁡〖x dx〗

    Since cos2x = 1 – sin2x, we have:

    ∫〖sin〗^n x dx= -cos⁡〖x 〖sin〗^(n-1) x+(n-1) ∫〖sin〗^(n-2) 〗 x dx-(n-1)∫〖sin〗^n x dx

    We solve this equation by taking the last term on the right side to the left side:

    ∫〖sin〗^n x dx= -cos⁡〖x 〖sin〗^(n-1) x+(n-1)∫〖〖sin〗^(n-2) x dx〗〗

    Is this right?
  2. jcsd
  3. Apr 17, 2008 #2
    Differentiate your answer using the same terms, see what you end up with.
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