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Thread: Calculus - Integration

  1. #1
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    Calculus - Integration

    how would I go about integrating (cos x)^3?
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    $\displaystyle \cos^3{x}= \cos^2{x}\cos{x}$
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    Quote Originally Posted by 11rdc11 View Post
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    $\displaystyle \cos^3{x}= \cos^2{x}\cos{x}$
    Alternatively

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

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

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


    Therefore

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

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

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