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Thread: Solve the equaton

  1. #1
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    Solve the equaton

    $\displaystyle

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

    \int\limits_{-\pi/2}^{\pi/2} sin^2x cos^2x (sinx+cosx)dx.
    $
    $\displaystyle \sin^2{x} \cos^2{x} (\sin{x}+\cos{x}) =$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    u = cosx ; du = - sinxdx

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

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

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

    Similarly


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

    v = sinx ; dv = cosxdx

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

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

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

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

    = $\displaystyle - \frac{4}{15}$...............Is this correct????
    Last edited by zorro; Dec 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 $\displaystyle - sinx$??????
    digging up the old posts, aren't you?

    anyway ...

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

    $\displaystyle (\cos^2{x} - \cos^4{x})\sin{x} = (\cos^4{x} - \cos^2{x})(-\sin{x})
    $
    mite but is my work correct ????
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