The correct option is A 2
I=∫sec θ(sec2θ)d θ
=sec θ ∫sec2θ dθ −∫tanθ (sec θtanθ)d θ
=sec θtanθ−∫sec θtan2θdθ
=sec θ tan θ−∫(sec2 θ−1)secθ d θ
=sec θ tan θ−∫ sec3 θ d θ+∫ sec θ d θ
=sec θ tan θ−I+ln|sec θ+tan θ|+C1
I=12[sec θ tan θ+ln|sec θ+tan θ|]+C
on comparing
a=2