Math integrals
posted by Marissa .
What is the indefinite integral of ∫ [sin (π/x)]/ x^2] dx ?
This is what I did:
Let u = π/x
(to get the derivative and du:)
π*1/x
π(1/x^2)dx = du
π(1/x^2)dx = (1)du
so, 1/x^2 = 1/πdu
then ∫ [sin (π/x)]/ x^2] dx = ∫ sin(u) (1/π)du
= 1/π ∫ sin u du
= 1/π (cos u) + C
= 1/π (cos u) + C
sub back in:
1/π cos (π/x) + C
I'm unsure of this because I don't know if I got du the right way.
Should it have been
Let u = π/x
π/x^2 dx = du
1/x^2 dx = πdu
That would make the answer a lot different...
Help!

Your first derivation is correct.
The last equation you wrote does not follow from (π/x^2) dx = du
You did it right the first time