Definite Integral

Question

!!int_0^(1/3) (ln(1+3x))/(1+9x^2) dx!!

(a) !!(pi/8)ln2!!

(b) !!(pi/12) ln2!!

(c) !!(pi/16) ln2!!

(d) !!(pi/24) ln2!!

Solution

!!I = int_0^(1/3) (ln(1+3x))/(1+9x^2) dx!!

Put !!3x= tantheta => 3dx= sec^2 theata!!

!!I = 1/3 int_0^(pi/4) (ln(1+tantheta))/(sec^2theta) xx sec^2 theta dtheta!!

!!=> I = 1/3 int_0^(pi/4)ln(1+tantheta) dtheta ....(1)!!

Use the property: !!int_a^b f(x) dx =int_a^b f(a+b-x)dx!!

!!=> I = 1/3 int_0^(pi/4)ln(1+tan(pi/4-theta) dtheta !!

!!=> I = 1/3 int_0^(pi/4)ln(1+ (1 - tantheta)/(1+tantheta))dtheta !!

!!=> I = 1/3 int_0^(pi/4)ln(2/(1+tantheta)) dtheta.....(2)!!

Adding !!(1)!! and !!(2)!!

!!2I = 1/3 int_0^(pi/4) ln(2)dtheta!!

!!=> I = 1/6(pi/4-0)ln2 !!

!!=> I = pi/24 ln2!!

Therefore, option (d) is correct.