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Sin[x]sin[2x]x dx again.

  1. Nov 22, 2003 #1
    [tex]\int^a_b x\sin{x}\sin{2x} dx[/tex]

    answers? I tried to solve once, it took like 3 pages.
     
  2. jcsd
  3. Nov 23, 2003 #2

    Hurkyl

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    To solve the integral [tex]\mbox{$\int x\sin x \sin 2x \,dx$}[/tex], let's first solve [tex]\mbox{$\int \sin x \sin 2x \,dx$.}[/tex]

    [tex]
    \begin{equation*}
    \begin{split}
    \int \sin{x}\sin{2x}
    &= \int 2 \sin^2 x \cos x\,dx \\
    &= \int 2 u^2 du \quad (u = \sin x) \\
    &= \frac{2}{3} u^3 + C\\
    &= \frac{2}{3} \sin^3 x + C
    \end{split}
    \end{equation*}
    [/tex]

    Now, to solve [tex]\mbox{$\int x\sin x \sin 2x \,dx$}[/tex] we may now do integration by parts via

    [tex]
    \begin{equation*}
    \begin{split}
    u &= x \\
    dv &= \sin x \sin 2x \,dx \\
    du &= dx \\
    v &= \frac{2}{3} \sin^3 x
    \end{split}
    \end{equation*}
    [/tex]

    Which yields:

    [tex]
    \begin{equation*}
    \begin{split}
    \int x\sin x \sin 2x \,dx
    &= x \frac{2}{3} \sin^3 x - \int \frac{2}{3} \sin^3 x \,dx \\
    &= \frac{2}{3} x \sin^3 x - \frac{2}{3} \int \sin x (1 - \cos^2 x)\,dx \\
    &= \frac{2}{3} x \sin^3 x - \frac{2}{3} \left( \int \sin x \, dx - \int \sin x \cos^2 x \, dx \right) \\
    &= \frac{2}{3} x \sin^3 x - \frac{2}{3}\left( -\cos x + \frac{1}{3} \cos^3 x \right) + C \\
    &= \frac{2}{3} x \sin^3 x + \frac{2}{3} \cos x - \frac{2}{9} \cos^3 x + C
    \end{split}
    \end{equation*}
    [/tex]
     
    Last edited: Nov 23, 2003
  4. Nov 23, 2003 #3
    Amazing. I have learned yet a new way from you.

    Out of curiousity, what background art thou?
     
  5. Nov 23, 2003 #4

    Hurkyl

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    Certainly not a professional TeX coding background. :smile: Took me what? Half an hour to get it that way? And even with that I couldn't get a nice table for the IBP. :frown:
     
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