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Thread: Final integrals

  1. #1
    Junior Member Selim's Avatar
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    Final integrals

    $\displaystyle \int sin(x)^4 * cos(x)^3 * dx$

    and

    $\displaystyle \int sin(x)^2 * cos(x) ^2 * dx$
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  2. #2
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    Quote Originally Posted by Selim View Post
    $\displaystyle \int sin(x)^4 * cos(x)^3 * dx$

    and

    $\displaystyle \int sin(x)^2 * cos(x) ^2 * dx$
    1. $\displaystyle \int{\sin^4{x}\cos^3{x}\,dx} = \int{\sin^4{x}\cos^2{x}\cos{x}\,dx}$

    $\displaystyle = \int{\sin^4{x}(1 - \sin^2{x})\cos{x}\,dx}$

    $\displaystyle = \int{\sin^4{x}\cos{x}\,dx} - \int{\sin^6{x}\cos{x}\,dx}$.

    Now in each integral, make the substitution $\displaystyle u = \sin{x}$.
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  3. #3
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    Quote Originally Posted by Selim View Post
    $\displaystyle \int sin(x)^4 * cos(x)^3 * dx$

    and

    $\displaystyle \int sin(x)^2 * cos(x) ^2 * dx$

    $\displaystyle \sin^4{x} \cdot \cos^3{x} =$

    $\displaystyle \sin^4{x} \cdot \cos^2{x} \cdot \cos{x} =
    $

    $\displaystyle \sin^4{x}(1-\sin^2{x}) \cdot \cos{x} =$

    $\displaystyle (\sin^4{x} - \sin^6{x}) \cdot \cos{x}$

    substitution ... let $\displaystyle u = \sin{x}$ , $\displaystyle du = \cos{x} \, dx$



    $\displaystyle \sin^2{x} \cdot \cos^2{x} =$

    $\displaystyle \left[\frac{1-\cos(2x)}{2}\right] \cdot \left[\frac{1+\cos(2x)}{2}\right] =
    $

    $\displaystyle \frac{1}{4} \left[1 - \cos^2(2x)\right] =$

    $\displaystyle \frac{1}{4} \left[\sin^2(2x)\right] =$

    $\displaystyle \frac{1}{4} \left[\frac{1 - \cos(4x)}{2}\right] =$

    $\displaystyle \frac{1}{8} \left[1 - \cos(4x)\right]$
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