dx 1- ( x 2 ) = (1+ x 2 ) (1- x 2 ) (1+ x 2 ) dx = ( 1+ x 2 1- ^2 x 2 )dx = ( 1+ x 2 ^2 x 2 )dx = ( ^2 x 2 + x 2 x 2 )dx = ( x 2 ) 1 2 + ( x 2 ) 1 2 +c =2 ( x 2 + x 2 )+c

1 1- x 2 dx

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$\int\frac{1}{1-\sin\frac{\text{x}}{2}}\text{dx}$

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Answer

$\int\frac{\text{dx}}{1-\sin\big(\frac{\text{x}}{2}\big)}$
$=\int\frac{\big(1+\sin\frac{\text{x}}{2}\big)}{\big(1-\sin\frac{\text{x}}{2}\big)\big(1+\sin\frac{\text{x}}{2}\big)}\text{dx}$
$=\int\bigg(\frac{1+\sin\frac{\text{x}}{2}}{1-\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\bigg(\frac{1+\sin\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\big(\sec^2\frac{\text{x}}{2}+\sec\frac{\text{x}}{2}\tan\frac{\text{x}}{2}\big)\text{dx}$
$=\frac{\tan\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}+\frac{\sec\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}+\text{c}$
$=2\big(\tan\frac{\text{x}}{2}+\sec\frac{\text{x}}{2}\big)+\text{c}$

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