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Find the integrals of the functions\[\sin x\;\sin2x\;\sin3x\]

$\begin{array}{1 1} \frac{[\frac{1}{3}\cos 6x-\frac{1}{2}\cos 4x-\cos 2x]}{8}+c. \\\large \frac{[\frac{1}{3}\cos 6x+\frac{1}{2}\cos 4x-\cos 2x]}{8}+c. \\ \large \frac{[\frac{1}{3}\cos 6x-\frac{1}{2}\cos 4x+\cos 2x]}{8}+c. \\ \frac{[\frac{1}{3}\cos 6x+\frac{1}{2}\cos 4x+\cos 2x]}{8}+c.\end{array} $

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  • (i)$\sin x.\sin y=\cos(x-y)-\cos (2x+3y).$
  • (ii)$2\sin x\cos y=\sin(x+y)+\sin(x-y).$
  • (iii)$\int \sin xdx=-\cos x+c$.
  • (iv)$\int \cos xdx=\sin x+c.$
Given $I=\int \sin x.\sin 2x.\sin 3xdx.$
Using the information in the tool box,we can write as,
$I=\int \sin x[\frac{1}{2}\cos(2x-3x)-\cos(2x+3x)]dx.$
$\;\;\;=\frac{1}{2}\int \sin x[\cos x-\cos 5x]dx.$
On multiplying we get,
$\;\;\;=\frac{1}{2}\int\sin x\cos xdx-\int \sin x\cos xdx.$
But $2\sin x\cos x=\sin2x$,hence
$\;\;\;=\frac{1}{2}[\int \frac{\sin 2x}{2}dx]-\frac{1}{2}\int \frac{1}{2}(\sin(x+5x)+\sin(x-5x)dx)$.
$\;\;\;=\frac{1}{4}\int\sin2xdx-\frac{1}{4}\int\sin6x-\sin 4xdx.$
On integrating we get,
$\;\;\;=\frac{1}{4}.\frac{1}{2}(-\cos 2x)-\frac{1}{4}[\frac{1}{6}(-\cos 6x)+\frac{1}{4}\cos 4x]+c.$
On multiplying we get,
$\;\;\;=\frac{-\cos2x}{8}+\frac{\cos 6x}{24}-\frac{1}{16}\cos 4x+c.$
Take $\frac{1}{8}$ as the common factor
$\;\;\;=\frac{1}{8}[\frac{1}{3}\cos 6x-\frac{1}{2}\cos 4x-\cos 2x]+c.$


answered Jan 30, 2013 by sreemathi.v
edited Jan 30, 2013 by sreemathi.v
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