Question

\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]

Answer 1

\[\text{ Let I }  = \int \frac{\cos x}{\cos 3x}dx\]
\[ = \int\frac{\cos x}{\left( 4 \cos^3 x - 3 \cos x \right)}dx \left[ \cos 3A = 4 \cos^3 A - 3 \cos A \right]\]
\[ = \int \frac{1}{4 \cos^2 x - 3}dx\]
\[\text{Dividing numerator and denominator by} \cos^2 x\]
\[ \Rightarrow I = \int \frac{\sec^2 x}{4 - 3 \sec^2 x} dx\]
\[ = \int \frac{\sec^2 x}{4 - 3\left( 1 + \tan^2 x \right)} dx\]
\[ = \int \frac{\sec^2 x}{1 - 3 \tan^2 x} dx\]
\[ = \int \frac{\sec^2 x}{1 - \left( \sqrt{3} \tan x \right)^2} dx\]
\[\text{ Let }\sqrt{3} \tan x = t\]
\[ \Rightarrow \sqrt{3} \sec^2 x \text{ dx }= dt\]
\[ \Rightarrow \sec^2 x \text{ dx } = \frac{dt}{\sqrt{3}}\]
\[ \therefore I = \frac{1}{\sqrt{3}} \int \frac{dt}{1^2 - t^2}\]
\[ = \frac{1}{\sqrt{3}} \times \frac{1}{2}\text{ ln } \left| \frac{1 + t}{1 - t} \right| + C\]
\[ = \frac{1}{2\sqrt{3}}\text{ ln } \left| \frac{1 + \sqrt{3} \tan x}{1 - \sqrt{3} \tan x} \right| + C\]