Question

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

Answer 1

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)]`)

Answer 2

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`