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Find the integrals of the functions\[\frac{\cos x}{1+\cos x}\]

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

1 Answer

Toolbox:
  • $1+\cos x=2cos^2\frac{x}{2}.$
  • $\cos x=2\cos^2\frac{x}{2}-1.$
  • $\int\sec^2x=\tan x+c.$
Given:$I=\int \frac{\cos x}{1+\cos x}$
 
Using the information from the tool box we get,
 
$\cos x=2\cos^2\frac{x}{2}-1$ and $1+\cos x=2\cos^2\frac{x}{2}.$
 
$I=\int\frac{2\cos^2\frac{x}{2}-1}{2\cos^2\frac{x}{2}}dx.$
 
Now seperating the terms we get,
 
$I=\int\big(1-\frac{1}{2\cos^2\frac{x}{2}}\big)dx.$
 
separating the terms we get,
 
$\;\;\;=\int dx-\frac{1}{2}\int \sec^2\frac{x}{2}dx.$
 
On integrating we get,
 
$\;\;\;=x-\frac{1}{2}\frac{\tan\frac{x}{2}}{\frac{1}{2}}+c.$
 
$\;\;\;=x-\tan x/2+c.$

 

answered Jan 30, 2013 by sreemathi.v
 
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