Question

\[\int\frac{\sin x}{\sqrt{4 \cos^2 x - 1}} dx\]

Answer 1

` ∫   { sin  x  dx }/{\sqrt{4 + cos^2 x -1}} `
\[\text{ let }\cos x = t\]
\[ \Rightarrow - \text{ sin x dx }= dt\]
\[ \Rightarrow \text{ sin x dx } = - dt\]
Now, ` ∫   { sin  x  dx }/{\sqrt{4 + cos^2 x -1}} `
\[ = \int\frac{- dt}{\sqrt{4 t^2 - 1}}\]
\[ = \int\frac{- dt}{\sqrt{4\left( t^2 - \frac{1}{4} \right)}}\]
\[ = - \frac{1}{2}\int\frac{dt}{\sqrt{t^2 - \left( \frac{1}{2} \right)^2}}\]
\[ = - \frac{1}{2} \text{ log }\left| t + \sqrt{t^2 - \frac{1}{4}} \right| + C\]
\[ = - \frac{1}{2} \text{ log } \left| t + \frac{\sqrt{4 t^2 - 1}}{2} \right| + C\]
\[ = - \frac{1}{2} \text{ log }\left| \frac{2t + \sqrt{4 t^2 - 1}}{2} \right| + C\]
\[ = - \frac{1}{2}\left[ \text{ log }\left| 2t + \sqrt{4 t^2 - 1} \right| - \text{ log 2 }  \right] + C\]
\[ = - \frac{1}{2} \text{ log }\left| 2t + \sqrt{4 t^2 - 1} \right| + \frac{\text{ log }  2}{2} + C\]
`  \text{ let C '} = {\log 2}/{2} + C `
` = -{1}/{2} log  |2 cos t + \sqrt{4 \cos^2 t - 1} \| + C `