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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\cos^42x\]

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Toolbox:
  • $(i)\sin^2x=\frac{(1-\cos 2x)}{2}.$
  • $(ii)\int\cos ^2xdx=\frac{(1+\cos2 x)}{2}+c.$
Given:$I=\int\cos^42xdx.$
 
But $\cos^22x=\frac{1+\cos 4x}{2}.$
 
This can be written as:
 
$I=\int(\cos^22x)^2dx=\int\bigg(\frac{1+\cos 4x}{2}\bigg)^2dx.$
 
Taking $\frac{1}{4}$ as the common factor
 
$\;\;\;=\frac{1}{4}\int(1+\cos 4x)^2dx.$
 
On separating the terms
 
$\;\;\;=\frac{1}{4}\int(1+\cos^24x-2\cos4x)dx.$
 
But $\cos^24x=\frac{(1+\cos 8x)}{2}$.
 
$\;\;\;=\frac{1}{4}\int[1+\bigg(\frac{1+\cos 8x}{2}\bigg)-2\cos 4x]dx.$
 
On separating the terms we get,
 
$I=\frac{1}{4}\int[\frac{3}{2}+\frac{1}{2}\cos 8x+\frac{1}{2}\cos 4x]dx.$
 
On integrating we get,
 
$I=\frac{3}{8}x+\frac{1}{8}\frac{1}{8}\sin 8x+\frac{1}{8}\sin 4x+c.$
 
On multiplying the factors
 
$\;\;\;=\frac{3}{8}x+\frac{\sin 8x}{64}+\frac{1}{8}\sin 4x+c.$
 
 

 

answered Jan 31, 2013 by sreemathi.v
 
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