what is the integral of sin³6x square root of cos6x?

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  • Is that sin(6x)^3 * sqrt(cos(6x)) * dx?

    u = cos(6x)

    du = -6 * sin(6x) * dx

    (1/6) * du = -sin(6x) * dx

    sin(6x)^3 * sqrt(cos(6x)) * dx =>

    sin(6x)^2 * sqrt(cos(6x)) * sin(6x) * dx =>

    (1 - cos(6x)^2) * sqrt(cos(6x)) * sin(6x) * dx =>

    (cos(6x)^2 - 1) * sqrt(cos(6x)) * (-sin(6x) * dx) =>

    (u^2 - 1) * sqrt(u) * (1/6) * du =>

    (1/6) * (u^(5/2) - u^(1/2)) * du

    Integrate

    (1/6) * ((2/7) * u^(7/2) - (2/3) * u^(3/2)) + C =>

    (1/6) * (1/7) * (1/3) * 2 * u^(3/2) * (3 * u^(4/2) - 7 * 1) + C =>

    (1/3) * (1/3) * (1/7) * (3u^2 - 7) * u^(3/2) + C =>

    (1/63) * (3 * cos(6x)^2 - 7) * cos(6x)^(3/2) + C

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  • 1 month ago

    You have sin(6x)*sin^2(6x)*sqrt[cos(6x)] dx.

    Let u = cos(6x), so (-1/6)du = sin(6x) dx.

    So now your integrand can be written as

    (-1/6)(1 - u^2)*sqrt(u)*du

    = (-1/6)[u^(1/2) - u^(5/2)] du.

    After integration you have

    (-1/6)[(2/3)u^(3/2) - (2/7)u^(7/2)] + C

    = (1/21)[cos(6x)]^(7/2) - (1/9)[cos(6x)]^(3/2) + C.

    check my arithmetic

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  • 1 month ago

    hint:wouldn't it be nice if you had sin to the power one? so that you could substitute u = cos 6x

    further hint - reduce the power of sin, by applying an identity.

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