How do you evaluate #int 1/(1+4x^2)# from #[0,sqrt3/2]#?

2 Answers
Nov 24, 2016

You integrate by reconducing to the fundamental integral with arcotangent primitive

Explanation:

#int_0^(sqrt(3)/2) dx / (1+4x^2) = int_0^(sqrt(3)/2) dx / (1 + (2x)^2) = 1/2 int_0^(sqrt(3)) dt / (1+t^2) = 1/2 arctan(t)|_0^(sqrt(3)) = 1/2 (arctan (sqrt(3)) - arctan(0)) = pi/6#

by substituting #t = 2x # and therefore #dx = dt /2#

(Remember when doing a substitution also the interval changes, so if x is in #[0,sqrt(3)/2]#, t is in #[0,sqrt(3)]#)

Nov 24, 2016

The answer is #=pi/6#

Explanation:

We evaluate this integral by substitution

Let #u=2x#

#du=2dx#

#intdx/(1+4x^2)=1/2int(du)/(1+u^2)#

Let #u=tanv#

#1+u^2=1+tan^2v=1/cos^2v#

#du=(dv)/cos^2v#

#1/2int(du)/(1+u^2)=1/2int(dv)/(cos^2v*1/cos^2v)#

#=1/2v=1/2*arctanu=1/2arctan2x+C#

#int_0 ^(sqrt3/2)dx/(1+4x^2)= [1/2*arctan2x ] _0^(sqrt3/2)#

#=1/2*arctansqrt3=1/2*pi/3=pi/6#