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# integration of 1/1+cosx , 1/1 -cos x

We can easily find the integration of $\frac {1}{1+ cos x}$ and $\frac {1}{1 -cos x}$ using trigonometric identities and fundamental integration techniques, The formula of these integral is given as

$\int \frac {1}{1+ cos x} \; dx = \tan \frac {x}{2} + C$

or

$\int \frac {1}{1+ cos x} \; dx = -\cot x + \csx x + C$

$\int \frac {1}{1- cos x} \; dx == -\cot {x}{2} + C$

or

$\int \frac {1}{1- cos x} \; dx = -\cot x -\csc x + C$

## Proof of integration of 1/1+cosx

Method I

$\int \frac {1}{1+ cos x} \; dx = \int \frac {1}{2\cos^2 \frac {x}{2} }$

$=\frac {1}{2} \int \sec^2 {x}{2} \; dx$

Now let ${x}{2} =t$

Then $dx= 2dt$
Therefore

$=\frac {2 }{2} \int \sec^2 t \; dt$

$=\tan t + C$
Substituting back

$= \tan {x}{2} + C$

Method II

$\int \frac {1}{1+ cos x} \; dx = \int \frac {1}{1+ cos x} \times \frac {1- cos x}{1- cos x}$

$= \int \frac {1- cos x}{1- cos^2 x } \; dx = \int \frac {1-cos x}{\sin^2 x} \; dx$

$=\int ( \csc^2 x – \cot x \csc x) \; dx$
$= -\cot x + \csx x + C$

## Proof of integration of 1/1-cosx

Method I

$\int \frac {1}{1- cos x} \; dx = \int \frac {1}{2\sin^2 \frac {x}{2} }$

$=\frac {1}{2} \int \csc^2 {x}{2} \; dx$

Now let ${x}{2} =t$

Then $dx= 2dt$
Therefore

$=\frac {2 }{2} \int \csc^2 t \; dt$

$=-\cot t + C$
Substituting back

$= -\cot {x}{2} + C$

Method II

$\int \frac {1}{1- cos x} \; dx = \int \frac {1}{1- cos x} \times \frac {1+ cos x}{1+ cos x}$

$= \int \frac {1+ cos x}{1- cos^2 x } \; dx = \int \frac {1+cos x}{\sin^2 x} \; dx$

$=\int ( \csc^2 x + \cot x \csc x) \; dx$
$= -\cot x – \csx x + C$

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