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Find the integrals of the functions\[\sin4x\;\sin8x\]

1 Answer

Toolbox:
  • (i)$2\sin x\sin y=\cos(A-B)-\cos(A+B).$
  • $(ii)\int\sin xdx=-\cos x+c.$
  • $(iii)\int\cos xdx=\sin x+c.$
Given $I=\int \sin 4x\sin 8x.$
 
Using the information from the tool box we write as,
 
$I=\frac{1}{2}\int[\cos(4x-8x)-\cos(4x+8x)]dx.$
 
$\;\;\;=\frac{1}{2}\int (\cos 4x-\cos 12x)dx.$
 
On separating the terms
 
$\;\;\;=\frac{1}{2}\int\cos 4xdx-\frac{1}{2}\int\cos 12x dx.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{2}[\frac{1}{4}\sin 4x]-\frac{1}{2}[\frac{1}{12}\sin 12x]+c.$
 
Taking $\frac{1}{4}$ as the common factor we get,
 
$\;\;\;=\frac{1}{4}\left\{[\frac{1}{2}\sin4x]-[\frac{1}{3}\sin 12x]\right\}+c$

 

answered Jan 30, 2013 by sreemathi.v
 
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