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Math Help - trigonometric integrals

  1. #1
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    trigonometric integrals

    Hi the problem is to evaluate the following

    integral sign ("S") b=pi/3 a=0 sin^3 x dx

    so using the identity 1= sin^2 x + cos^2 x

    S sin^3 x dx = S sin x (1-cos^2 x) dx

    So if I let u= cos x du = sin x dx

    Then the original equation becomes = S (1-u^2) du
    = u - (1/3) u^3
    Substituting back and now applying a= 0 and b = pi/3
    = (cos pi/3 - (1/3) cos^3 pi/3) - (cos 0 - (1/3) cos^3 0)

    First am I doing this right?? And second if this is right does cos^3 pi/3 = 1/8?

    Thanks
    Calculus Beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    Hi the problem is to evaluate the following

    integral sign ("S") b=pi/3 a=0 sin^3 x dx

    so using the identity 1= sin^2 x + cos^2 x

    S sin^3 x dx = S sin x (1-cos^2 x) dx

    So if I let u= cos x du = sin x dx ... du = -sin x dx

    ...
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  3. #3
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    now regarding cos^3 pi/3

    is cos^3 pi/3 = (1/2)^3 = 1/8???

    Thanks

    Calculus Beginner
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  4. #4
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    Quote Originally Posted by calcbeg View Post
    is cos^3 pi/3 = (1/2)^3 = 1/8???

    Thanks

    Calculus Beginner
    yes
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