Question

\[\int\frac{\sin x}{\sqrt{\cos^2 x - 2 \cos x - 3}} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\frac{\sin x}{\sqrt{\cos^2 x - 2 \cos x - 3}}dx\]
\[\text{ Putting  cos  x  = t}\]
\[ \Rightarrow - \text{ sin  x  dx }= dt\]
\[ \Rightarrow \text{ sin  x  dx } = - dt\]
\[ \therefore I = - \int\frac{dt}{\sqrt{t^2 - 2t - 3}}\]
\[ = - \int\frac{dt}{\sqrt{t^2 - 2t + 1 - 4}}\]
\[ = - \int\frac{dt}{\sqrt{\left( t - 1 \right)^2 - \left( 2 \right)^2}}\]
\[ = - \text{ ln }\left| t - 1 + \sqrt{\left( t - 1 \right)^2 - 4} \right| + C ..........................\left[ \because \int\frac{1}{\sqrt{x^2 - a^2}}dx = \text{ ln}\left| x + \sqrt{x^2 - a^2} \right| + C \right]\]
\[ = - \text{ ln }\left| \left( \cos x - 1 \right) + \sqrt{\cos^2 x - 2 \cos x - 3} \right| + C.................... \left[ \because t = \cos x \right]\]