Integral of cot(x)csc^2(x)

Question

I know how to do this problem correctly setting u = cot(x)

but at first I tried setting u = sin(x) then I converted the whole cot(x)csc^2(x) into (cosx/sinx)(1/sin^2(x)). u = sin(x) and du = cos(x) so plug it in and you should get du/u^3.

Then after doing the antiderivative I got -1/2 * u^-2 which is -1/2sin^2(x) + C

Did I do something wrong?

Answer 1

No, the answers are correct. In the first part, you get -1/2 cot²x + C₁ and in the second part, you get -1/2 csc²x + C₂ .(or -1/2 sin^-2x + C₂) Note that cot² + 1 = csc² so they are connected by a constant that can be absorbed into the general constant. If you have a definite integral, then the constants would make the two expressions equal.