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Math Help - Solve the equaton

  1. #1
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    Solve the equaton

    <br /> <br />
\int\limits_{-\pi/2}^{\pi/2} sin^2x  cos^2x (sinx+cosx)dx.<br />
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  2. #2
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    Quote Originally Posted by zorro View Post
    <br /> <br />
\int\limits_{-\pi/2}^{\pi/2} sin^2x  cos^2x (sinx+cosx)dx.<br />
    \sin^2{x}  \cos^2{x} (\sin{x}+\cos{x}) =

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

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

    (\cos^2{x} - \cos^4{x})\sin{x} + (\sin^2{x} - \sin^4{x})\cos{x}

    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^4{x} - \cos^2{x})(-\sin{x}) \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\sin^2{x} - \sin^4{x})\cos{x} \, dx

    for the first integral, let u = \cos{x}

    for the second integral, let u = \sin{x}
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \sin^2{x}  \cos^2{x} (\sin{x}+\cos{x}) =

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

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

    (\cos^2{x} - \cos^4{x})\sin{x} + (\sin^2{x} - \sin^4{x})\cos{x}

    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^4{x} - \cos^2{x})(-\sin{x}) \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\sin^2{x} - \sin^4{x})\cos{x} \, dx

    for the first integral, let u = \cos{x}

    for the second integral, let u = \sin{x}
    how did u get a - sinx??????
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  4. #4
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    Quote Originally Posted by skeeter View Post
    \sin^2{x}  \cos^2{x} (\sin{x}+\cos{x}) =

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

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

    (\cos^2{x} - \cos^4{x})\sin{x} + (\sin^2{x} - \sin^4{x})\cos{x}

    \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^4{x} - \cos^2{x})(-\sin{x}) \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\sin^2{x} - \sin^4{x})\cos{x} \, dx

    for the first integral, let u = \cos{x}

    for the second integral, let u = \sin{x}
    I_1 = \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} (cos^4 x - cos^2 x)(- sinx)dx

    u = cosx ; du = - sinxdx

    = \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} (u^4  - u^2 )du

     \frac{1}{5} \left[ cos \left( - \frac{\pi}{2} \right) - cos \left( \frac{ \pi}{2} \right) \right]^5 - \frac{1}{3} \left[ cos \left( - \frac{\pi}{2} \right) - cos \left( \frac{ \pi}{2} \right) \right]^3

    = 0 ...........Is this correct ?????

    Similarly


    I_2 = \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} (sin^2 x - sin^4 x)(cosx)dx

    v = sinx ; dv = cosxdx

    = - \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} (u^2  - u^4 )du

    - \frac{1}{3} \left[ sin \left( - \frac{\pi}{2} \right) - sin \left( \frac{ \pi}{2} \right) \right]^3 - \frac{1}{5} \left[ sin \left( - \frac{\pi}{2} \right) - sin \left( \frac{ \pi}{2} \right) \right]^5

    =  \frac{1}{3} (-1-1) - \frac{1}{5} (-1-1)

    = -\frac{2}{3} + \frac{2}{5}

    = - \frac{4}{15}...............Is this correct????
    Last edited by zorro; December 27th 2009 at 11:27 PM.
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  5. #5
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    Quote Originally Posted by zorro View Post
    how did u get a - sinx??????
    digging up the old posts, aren't you?

    anyway ...

    (\cos^2{x} - \cos^4{x})\sin{x} = (\cos^4{x} - \cos^2{x})(-\sin{x})<br />
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  6. #6
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    Quote Originally Posted by skeeter View Post
    digging up the old posts, aren't you?

    anyway ...

    (\cos^2{x} - \cos^4{x})\sin{x} = (\cos^4{x} - \cos^2{x})(-\sin{x})<br />
    mite but is my work correct ????
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