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Posted by on Thursday, June 28, 2012 at 2:07pm.

(SIN^8X-5 SIN^5X+SINX)COSX DX

I need the answer to this, because I have tried so many times and cannot figure it out.

  • calculus - , Thursday, June 28, 2012 at 4:02pm

    I infer from the presence of DX that you want

    ∫(sin^8(x) - 5sin^5(x) + sin(x)) cos(x) dx

    if you let u = sin(x), then du = cos(x) dx you have

    ∫(u^8 - 5u^5 + u) du

    that's dead easy, using thge power rule:

    1/9 u^9 - 1/6 u^6 + 1/2 u^2 + C

    now change all the u's back to sin(x) and you're done

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