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# Integrate the function$x\;\sin3x$

Can you answer this question?

Toolbox:
• (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

du=dx.

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