Question

How do you find the integral of `int (1/(1+x^2) ) dx` from negative infinity to 0?

Answer 1

`pi/2`

Explanation:

let `x=tantheta`
`dx=sec^2theta d(theta)`
`1+tan^2theta=sec^2theta`
`if x=-oo theta=-pi/2`
`if x=9 theta=0`
`int_-oo^01/(1+x^2)dx=int_-(pi/2)^0sec^2theta/sec^2thetad(theta)`
=`int_-(pi/2)^0d(theta)`=`[theta]_(-pi/2)^0=0-(-pi/2)=pi/2`

Answer 2

This is related to `arctanx`...

`d/(dx)[arctanx] = 1/(1+x^2)`

`:. " " int_a^b 1/(1+x^2)dx = arctanx + C`

`int_(-oo)^(0) 1/(1+x^2) = arctan(0) - arctan(-oo)`

Due to `tanx` having a domain of `(pi/2, pi/2) pm pik` where `k in ZZ`, the inverse, `arctanx`, is a graph with a range of `(-pi/2,pi/2)`, and it looks kind of like a sideways `x^3`, crossing through `(0,0)`.

graph{arctanx [-20, 20, -3.12, 3.12]}

Therefore, as `x->-oo`, you get `-pi/2`, and so:

`= 0 - (-pi/2) = color(blue)(pi/2)`