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Integrate the function $\;\int\;x\;sin\;x$

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  • Where there are two functions u and v,and if they are of the form $\int u dv,$ then by the method of integration by parts we can solve them as,$\int udv=uv-\int vdu.$
  • (ii)$\int\sin xdx=-\cos x+c.$
  • $(iii)\int \cos xdx=\sin x+c.$
Given $I=\int x\sin xdx.$
It is very clean that the given problem is of the form $\int udv.$
We can follow the method of substitution to solve this
$\int u dv=uv-\int vdu.$
Let u=x.
On differentiating we get
Let du=dx.
Let $dv=\sin xdx.$
On substituting for u and dv we get
$\int x\sin xdx=[x.(-\cos x)]-\int-\cos xdx.$
On integrating we get
$\;\;\;=-x\cos x+\sin x+c.$
Hence $\int x\sin xdx=-x\cos x+\sin x+c.$


answered Feb 8, 2013 by sreemathi.v