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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\frac{1-\cos x}{1+\cos x}\]

$\begin{array}{1 1} 2\tan \frac{x}{2}-x+c. \\ 2\tan \frac{x}{2}+x+c \\ 2\cot \frac{x}{2}-x+c \\ 2\cot \frac{x}{2}+x+c. \end{array} $

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1 Answer

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Toolbox:
  • (i)$1-\cos x=2\sin^2\frac{x}{2}$.
  • (ii)$1+\cos x=2\cos^2\frac{x}{2}$.
  • (iii)$\int \tan xdx=log|\sec x|+c.$
  • (iv)$\sec^2xdx=\tan x+c.$
Given $I=\int\frac{1-\cos x}{1+\cos x}dx.$
 
Using the information in the tool box we get
 
$1-\cos x=2\sin^2\frac{x}{2}$ and $1+\cos x=2\cos^2\frac{x}{2}$.
 
$I=\int\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}=\int\tan^2\frac{x}{2}.$
 
But $\tan^2\frac{x}{2}=\sec^2\frac{x}{2}-1$.
 
Substituting this,
 
$I=\int(\sec^2\frac{x}{2}-1)dx.$
 
On separating the terms,
 
$\;\;\;=\int\sec^2\frac{x}{2}dx-\int dx.$
 
On integrating we get,
 
$\;\;\;=\frac{\tan\frac{x}{2}}{\frac{1}{2}}-x+c.$
 
$\;\;\;=2\tan\frac{x}{2}-x+c.$

 

answered Jan 30, 2013 by sreemathi.v
 
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