Question

\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]

Answer 1

\[\int\frac{1}{\sqrt{1 - \cos 2x}}dx\]
\[ = \int\frac{1}{\sqrt{2 \sin^2 x}}dx \left[ \because 1 - \cos 2x = 2 \ sin^2 x \right]\]

` = 1/sqrt2 ∫  "cosec"  x  dx `
\[ = \frac{1}{\sqrt{2}}\text{ln }\left| \text{cosec x} - \text{ cot x} \right| + C\] 

` =   1/\sqrt{2}  In  | 1/ sin x  -  cos x / sin x| + C`

` =   1/\sqrt{2}  In  | {2 sin ^{2 x/2}} / sin x | + C `     ` [ ∵  1 - cos x = 2   sin^2  x/2 ]`

 

` =   1/\sqrt{2}  In  |   {2 sin ^{2x/2}} / {2sin x/2  cos  x/2 } |` + C      ` [ ∵  sin x  = 2   sin  x/2      cos  x/2 ]`
\[ = \frac{1}{\sqrt{2}} \text{ln} \left| \tan\frac{x}{2} \right| + C\]