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Integrate the function\[x\;\sin3x\]

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  • (i)Let us consider two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts\[\int udv=uv-\int vdu\]
  • (ii)$\sin axdx=\frac{-1}{a}\cos ax+c.$
  • (iii)$\int\cos axdx=\frac{1}{a}\sin ax+c.$
Given $I=\int x\sin 3xdx.$
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts where \[\int udv=uv-\int vdu\]
Let u=x.
On differentiating we get
Let $dv=\sin 3xdx.$
On integrating on both sides we get
$v=\frac{-1}{3}\cos 3x.$
Hence on substituting for u.v.dv and du,we get
$\int x\sin 3xdx=(x)\big(\frac{-1}{3}\sin 3x\big)-\int\frac{1}{-3}\sin 3xdx.$
On multiplying the symbols we get
$\;\;\;=\big(\frac{-x}{3}\sin 3x\big)+\frac{1}{3}\int \cos 3xdx.$
$\;\;\;=\big(\frac{-x}{3}\sin 3x)+\frac{1}{3}\frac{\sin 3x}{3}+c.$
$\;\;\;=\frac{-x}{3}\sin 3x+\frac{1}{9}\sin 3x+c.$


answered Feb 8, 2013 by sreemathi.v