I must be missing something

  • Thread starter Bazzinga
  • Start date
  • #1
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So I've gotten the integral that i'm doing now down to:

int(cos(x)/sin^2(x) dx)

I looked it up on one of those online integral calculators to get me on the right track, and the answer is:

-1/sin(x)

It seems so simple, what am I missing?
 

Answers and Replies

  • #2
34,686
6,393
An ordinary substitution will do here: u = sin(x), du = cos(x)dx. Then your integral is
[tex]\int \frac{du}{u^2} = \int u^{-2}du[/tex]
 

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