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Math Help - Calculus - Integration

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    Calculus - Integration

    how would I go about integrating (cos x)^3?
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    Super Member 11rdc11's Avatar
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    \cos^3{x}= \cos^2{x}\cos{x}
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    Quote Originally Posted by 11rdc11 View Post
    change

    \cos^3{x}= \cos^2{x}\cos{x}
    Alternatively

    Since \cos{3x} = 4\cos^3{x} - 3\cos{x}

    4\cos^3{x} = \cos{3x} + 3\cos{x}

    \cos^3{x} = \frac{1}{4}\cos{3x} + \frac{3}{4}\cos{x}.


    Therefore

    \int{\cos^3{x}\,dx} = \int{\frac{1}{4}\cos{3x} + \frac{3}{4}\cos{x}\,dx}

     = \frac{1}{12}\sin{3x} + \frac{3}{4}\sin{x} + C.
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    Quote Originally Posted by 11rdc11 View Post
    change

    \cos^3{x}= \cos^2{x}\cos{x}
    @OP: Then replace \cos^2 x with 1 - \sin^2 x. Then make the substitution u = \sin x.
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