Question

\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]

Answer 1

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