Question

How do you find the indefinite integral of `int 4/(1+t^2)dt`?

Answer 1

`int 4/(1+t^2)dt =4arctant+C`

Explanation:

Substitute:

`t=tanx`

`dt = (dx)/(cos^2x)`

`x=arctan t`

So:

`int 4/(1+t^2)dt = 4 int 1/(1+tan^2x) (dx)/(cos^2x)=4 int 1/(1+sin^2x/cos^2x) (dx)/(cos^2x)=4 int (dx)/(cos^2x+sin^2x) =4 int dx = 4x+C = 4arctant+C`