Question

How do you integrate `int 1/theta^2cos(1/theta)`?

Answer 1

`-sin(1/theta) + C`

Explanation:

Let `u = 1/theta`. Then `du = -1/theta^2 d theta` and `d theta = -du(theta)^2`.

`=int(1/theta^2 cosu *-du (theta)^2) du`

`= -int(cosu)du`

`=-sinu + C`

Resubstitute:

`=-sin(1/theta) + C`

Hopefully this helps!