Question

\[\int\frac{1}{\cos 2x + 3 \sin^2 x} dx\]

Answer 1

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