Calculus

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Evaluate ∫ (cos(x))^(1/2)sin(x)dx

Let u = cos(x)?

∫ (u)^(1/2)sin(x)dx = ∫ [2u^(3/2)/3]sin(x)dx
∫ [2cos(x)^(3/2)/3] (-cos(x)) dx?

I thought this involved the FTC, but now I'm thinking that's false.

  • Calculus -

    ok, let u = cos x
    then du = -sin x dx

    then you have

    ∫ (u)^(1/2)(-du)

    =-(2/3) u^(3/2) + c

  • Calculus -

    of course that is

    -(2/3) (cos x)^3/2 + c

    Your problem seems to be not replacing dx with a calculated du

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