Question

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

Answer 1

\[\text{ Let I } = \int\frac{\cos x}{\frac{1}{4} - \cos^2 x}dx\]

\[ = \int\frac{\cos x}{\frac{1}{4} - \left( 1 - \sin^2 x \right)}dx\]

\[ = \int\frac{\cos x}{\sin^2 x - \frac{3}{4}}dx\]

\[ = \int\frac{\cos x}{\sin^2 x - \left( \frac{\sqrt{3}}{2} \right)^2}dx\]

\[\text{ Putting  sin x = t}\]

\[ \Rightarrow \text{ cos  x  dx = dt }\]

\[ \therefore I = \int\frac{1}{t^2 - \left( \frac{\sqrt{3}}{2} \right)^2}dt\]

\[ = \frac{1}{2 \times \frac{\sqrt{3}}{2}} \text{ ln  }\left| \frac{t - \frac{\sqrt{3}}{2}}{t + \frac{\sqrt{3}}{2}} \right| + C\]

\[ = \frac{1}{\sqrt{3}} \text{ ln } \left| \frac{2t - \sqrt{3}}{2t + \sqrt{3}} \right| + C\]

\[ = \frac{1}{\sqrt{3}} \text{ ln } \left| \frac{2 \sin x - \sqrt{3}}{2 \sin x + \sqrt{3}} \right| + C................ \left[ \because t = \sin x \right]\]