Posted by **Marissa** on Friday, August 15, 2008 at 1:31am.

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!

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