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Math Help - Final integrals

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

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

    and

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

    and

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

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

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

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

    and

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

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

    \sin^4{x} \cdot \cos^2{x} \cdot \cos{x} =<br />

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

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

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



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

    \left[\frac{1-\cos(2x)}{2}\right] \cdot \left[\frac{1+\cos(2x)}{2}\right] =<br />

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

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

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

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