Given:
The expression is:
[tex]\dfrac{1-\cos 2x}{1+\cos 2x}[/tex]
To find:
The integration of the given expression.
Solution:
We need to find the integration of [tex]\dfrac{1-\cos 2x}{1+\cos 2x}[/tex].
Let us consider,
[tex]I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx[/tex]
[tex]I=\int \dfrac{2\sin^2x}{2\cos^2x}dx[/tex] [tex][\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x][/tex]
[tex]I=\int \dfrac{\sin^2x}{\cos^2x}dx[/tex]
[tex]I=\int \tan^2xdx[/tex] [tex]\left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right][/tex]
It can be written as:
[tex]I=\int (\sec^2x-1)dx[/tex] [tex][\because 1+\tan^2 \theta =\sec^2 \theta][/tex]
[tex]I=\int \sec^2xdx-\int 1dx[/tex]
[tex]I=\tan x-x+C[/tex]
Therefore, the integration of [tex]\dfrac{1-\cos 2x}{1+\cos 2x}[/tex] is [tex]I=\tan x-x+C[/tex].
