Question

integration of arctan(secx+tanx)
x lying between -pi/2 to pi/2​

Answer 1

The problem is that you divide the numerator and the denominator of the integrand by cos2x, but this can be done only for cos2x≠0 i.e. x≠π/2. So you have to divide your integral in two improper integrals as:

∫π0=∫(π/2)−0−∫(π/2)+π

Your substutition is correct, so you have the limit:

limx→(π/2)±arctan(batanx)

that is ±π/2.

Answer 2

Answer:

I hope you are satisfied