Question

What is the definite integral of `1/(36+x^2)` with bounds `[0, 6]`?

Answer 1

`pi/24`

Explanation:

There is a standard integral: `int(1/(a^2+x^2))dx = 1/atan^-1(x/a)+c`

So `int_0^6(1/(36+x^2))dx = [1/6tan^-1(x/6)]_0^6`

`=(1/6tan^-1(1))-(1/6tan^-1(0)) = 1/6*pi/4 - 1/6*0 = pi/24`