Question

Find the integrals of the function: $\frac {cos 2x+2 sin^2x}{cos^2x}$

Answer 1

$\frac {cos 2x+2 sin^2x}{cos^2x}$
=$\frac {cos\ 2x+(1-cos \ 2x)}{cos^2x }$ [cos 2x = 1-2sin2x]
= $\frac {1}{cos^2 x}$
= sec2 x
∴ ∫$\frac {cos 2x+2 sin^2x}{cos^2x}$=∫sec2x dx = tan x +C