Question

\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]

Answer 1

\[\text{ Let I} = \int \frac{1}{4 \cos x - 1}dx\]
\[\text{ Putting  cos x }= \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]
\[ \Rightarrow I = \int \frac{1}{4\left( \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \right) - 1}dx\]
\[ = \int \frac{1}{\frac{4\left( 1 - \tan^2 \frac{x}{2} \right) - \left( 1 + \tan^2 \frac{x}{2} \right)}{\left( 1 + \tan^2 \frac{x}{2} \right)}}\]
\[ = \int \frac{\left( 1 + \tan^2 \frac{x}{2} \right)dx}{4 - 4 \tan^2 \left( \frac{x}{2} \right) - 1 - \tan^2 \left( \frac{x}{2} \right)}\]
\[ = \int \frac{\sec^2 \left( \frac{x}{2} \right) dx}{3 - 5 \tan^2 \left( \frac{x}{2} \right)}\]
\[\text{ Let tan } \left( \frac{x}{2} \right) = t\]
\[ \Rightarrow \frac{1}{2} \sec^2 \left( \frac{x}{2} \right)\text{ dx }= dt\]
\[ \Rightarrow \sec^2 \left( \frac{x}{2} \right)dx = 2dt\]
\[ \therefore I = 2 \int \frac{dt}{3 - 5 t^2}\]
\[ = \frac{2}{5} \int \frac{dt}{\frac{3}{5} - t^2}\]
\[ = \frac{2}{5} \int \frac{dt}{\left( \frac{\sqrt{3}}{\sqrt{5}} \right)^2 - t^2}\]
\[ = \frac{2}{5} \times \frac{\sqrt{5}}{2\sqrt{3}}\text{ In }\left| \frac{\frac{\sqrt{3}}{\sqrt{5}} + t}{\frac{\sqrt{3}}{\sqrt{5}} - t} \right| + C\]
\[ = \frac{1}{\sqrt{15}}\text{ ln } \left| \frac{\sqrt{3} + \sqrt{5} t}{\sqrt{3} - \sqrt{5} t} \right| + C\]
\[ = \frac{1}{\sqrt{15}}\text{  ln }\left| \frac{\sqrt{3} + \sqrt{5} \tan \left( \frac{x}{2} \right)}{\sqrt{3} - \sqrt{5} \tan \left( \frac{x}{2} \right)} \right| + C\]