The correct option is C None of these
Option A,
I=∫π/20ln(cotx)dx
I=∫π/20ln(tanx)dx
⇒2I=∫π/20ln(tanxcotx)dx
⇒2I=∫π/20ln(1)dx
⇒I=0
Option B,
I=∫2π0sin3xdx
I=14∫2π0(sin3x−3sinx)dx
I=14(−13cos3x−3cosx)2π0
⇒I=0
Option C, ∫e1/edxx(lnx)1/3
Put lnx=t
1xdx=dt
I=∫1−1dtt1/3
I=23[t2/3]1−1