Question

Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .

Answer 1

`(cos2x + 2sin^2 x)/ cos^2x`

= `(cos 2x + (1 - cos 2x))/cos^2`        [cos2x = 1-2sin² x]

=` 1/(cos2 x)`

= `sec^2 x`

∴ \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] = \[\int\]`sec^2x  dx = tan x +c`