#int_(1/a)^a(tan^-1x)/xdx=?#

Question

1628 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-22T09:08:01

Let
# z=lnx=>x=e^z#
# x=a->z=lna and x=1/a->z=-lna#

#and dz==(dx)/x#

#I=int_(1/a)^a(tan^-1x)/xdx#

#color(red)("Replacing "a" by "lna , 1/a" by "-lna,x" by "e^z and {dx}/x" by "dz)#

#I=int_(-lna)^(lna)tan^-1(e^z)dz#

#=int_(-lna)^0tan^-1e^zdz+int_0^(lna)tan^-1e^zdz#

#=int_0^(lna)tan^-1e^-zdz+int_0^(lna)tan^-1e^zdz#

#=int_0^(lna)(cot^-1e^z+tan^-1e^z)dz#

#color(blue)("Applying formula " tan^-1x+cot^-1x=pi/2)#

#=int_0^(lna)(pi/2)dz#

#=pi/2lna#