Question

\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]

Answer 1

\[\text{ Let I } = \int \frac{dx}{\sqrt{3} \sin x + \cos x}\]
\[\text{ Putting sin x } = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \text{ and cos x} = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]
\[ \Rightarrow I = \int \frac{1}{\sqrt{3}\frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}dx\]
\[ = \int \frac{1 + \tan^2 \frac{x}{2}}{2\sqrt{3} \tan \frac{x}{2} + 1 - \tan^2 \frac{x}{2}}dx\]
\[ = \int\frac{\sec^2 \frac{x}{2}}{- \tan^2 \frac{x}{2} + 2\sqrt{3} \tan \frac{x}{2} + 1}dx\]

\[\text{ Let tan} \frac{x}{2} = t\]
\[ \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2}dx = dt\]
\[ \Rightarrow \sec^2 \frac{x}{2}dx = 2dt\]
\[ \therefore I = 2\int \frac{dt}{- t^2 + 2\sqrt{3}t + 1}\]
\[ = - 2\int \frac{dt}{t^2 - 2\sqrt{3}t - 1}\]
\[ = - 2\int\frac{dt}{t^2 - 2\sqrt{3}t + \left( \sqrt{3} \right)^2 - \left( \sqrt{3} \right)^2 - 1}\]
\[ = - 2\int \frac{dt}{\left( t - \sqrt{3} \right)^2 - \left( 2 \right)^2}\]
\[ = - \frac{2}{2 \times 2}\text{ log }\left| \frac{t - \sqrt{3} - 2}{t - \sqrt{3} + 2} \right| + C\]

\[= - \frac{1}{2}\text{ log}\left| \frac{\tan\frac{x}{2} - 2 - \sqrt{3}}{\tan\frac{x}{2} + 2 - \sqrt{3}} \right| + C\]
\[ = \frac{1}{2}\text{ log}\left| \frac{\tan\frac{x}{2} + 2 - \sqrt{3}}{\tan\frac{x}{2} + 2 - \sqrt{3}} \right| + C\]