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Math Help - [SOLVED] evaulating the integral- integration by parts

  1. #1
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    Post [SOLVED] evaulating the integral- integration by parts

    \int_0^\infty  {x\sin (x)\cos (x)dx}<br />

    I tried:
    u=xsinx, du=xcox+sinx
    v=sinx, dv=cosx
    (x\sin x)(\sin x) - \int {(xcoxs + \sin x)(\sin x)dx} <br />
    (x\sin x)(\sin x) - \int {x\cos x\sin x + {{\sin }^2}xdx} <br />
    but then I got stuck on integrating...
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by genlovesmusic09 View Post
    \int_0^\infty {x\sin (x)\cos (x)dx}<br />

    I tried:
    u=xsinx, du=xcox+sinx
    v=sinx, dv=cosx
    (x\sin x)(\sin x) - \int {(xcoxs + \sin x)(\sin x)dx} <br />
    (x\sin x)(\sin x) - \int {x\cos x\sin x + {{\sin }^2}xdx} <br />
    but then I got stuck on integrating...
    Recall that 2sinxcosx = sin(2x).

    So I would recommend that you rewrite the integral as \tfrac{1}{2}\int_0^{\infty} x\sin\!\left(2x\right)\,dx
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  3. #3
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    Quote Originally Posted by genlovesmusic09 View Post
    \int_0^\infty  {x\sin (x)\cos (x)dx}<br />

    I tried:
    u=xsinx, du=xcox+sinx
    v=sinx, dv=cosx
    (x\sin x)(\sin x) - \int {(xcoxs + \sin x)(\sin x)dx} <br />
    (x\sin x)(\sin x) - \int {x\cos x\sin x + {{\sin }^2}xdx} <br />
    but then I got stuck on integrating...
    Another way:

    Since \int xsin(x)cos(x) = xsin^2(x) - \int xsin(x)cos(x) - \int sin^2(x),

    We get:

    2 \int xsin(x)cos(x) = xsin^2(x) - \int sin^2(x)

    From here it should be easy (find the RHS and divide it by 2 to get the answer).
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