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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\sin^4x\]

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Toolbox:
  • $(i)1-\sin^2x=\cos^2x.$
  • $(ii)\int\cos xdx=\sin x+c.$
  • $(iii)\sin^2x=\frac{(1-\cos 2x)}{2}.$
  • $(iv)\cos^2x=\frac{(1+\cos 2x)}{2}.$
Given:$I=\int\sin^4xdx.$
 
This can be written as,
 
$I=\int(\sin^2x)^2dx$.
 
But $\sin^2x=\frac{(1-\cos 2x)}{2}.$
 
$\;\;\;=\int\bigg(\frac{1-\cos 2x}{2}\bigg)\bigg(\frac{1-\cos 2x}{2}\bigg)dx.$
 
$\;\;\;=\frac{1}{4}\int(1-\cos2x)^2dx.$
 
This is the form $(a-b)^2=a^2-2ab+b^2.$
 
$\;\;\;=\frac{1}{4}\int(1+\cos^22x-2\cos2x)dx.$
 
But $\cos^22x=\frac{(1+\cos 4x)}{2}$.
 
$\;\;\;=\frac{1}{4}\int[1+\bigg(\frac{1+\cos 4x}{2}\bigg)-2\cos 2x]dx.$
 
On separating the terms we get,
 
$1+\frac{1}{2}=\frac{3}{2}.$
 
$I=\frac{1}{4}\int[\frac{3}{2}+\frac{1}{2}\cos 4x-2\cos 2x]dx.$
 
on separating terms we get,
 
$\;\;\;=\frac{3}{8}\int dx+\frac{1}{8}\int \cos 4xdx-\frac{1}{2}\cos 2xdx.$
 
On integrating we get,
 
$\;\;\;=\frac{3}{8}x+\frac{1}{8}\frac{1}{4}\sin 4x-\frac{1}{2}\frac{1}{2}\sin 4x+c.$
 
On multiply
 
$\;\;\;=\frac{3}{8}x+\frac{1}{32}\sin 4x-\frac{1}{4}\sin 4x+c.$
 
Taking $\frac{1}{8}$ as a common factor
 
$\;\;\;=\frac{1}{8}[3x+\frac{1}{4}\sin 4x-2\sin 4x]+c.$

 

answered Jan 30, 2013 by sreemathi.v
 
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