( 1+ 1- )dx = ( 2 ^2 x 2 2 ^2 x 2 )dx [ 1+ =2 ^2 x 2 &1- =2 ^2 x 2 ] = ^2 x 2 dx = (cosec^2 x 2 -1 )dx = - ( x 2 ) 1 2 -x+c =-2 ( x 2 )-x+c

1+ 1- dx

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$\int\frac{1+\cos\text{x}}{1-\cos\text{x}}\text{dx}$

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Answer

$\int\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\text{dx}$
$=\int\bigg(\frac{2\cos^2\frac{\text{x}}{2}}{2\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$ $\big[\therefore1+\cos\text{x}=2\cos^2\frac{\text{x}}{2} \&1-\cos\text{x}=2\sin^2\frac{\text{x}}{2}\big]$
$=\int\cot^2\frac{\text{x}}{2}\text{dx}$
$=\int\Big(\text{cosec}^2\frac{\text{x}}{2}-1\Big)\text{dx}$
$=\frac{-\cot\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}-\text{x+c}$
$=-2\cot\big(\frac{\text{x}}{2}\big)-\text{x+c}$

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