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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} $

1 Answer

  • (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 $1+\cos x=2\cos^2\frac{x}{2}$.
But $\tan^2\frac{x}{2}=\sec^2\frac{x}{2}-1$.
Substituting this,
On separating the terms,
$\;\;\;=\int\sec^2\frac{x}{2}dx-\int dx.$
On integrating we get,


answered Jan 30, 2013 by sreemathi.v