How do you find the antiderivative of $\cot^2x$?
My steps to find it
First
$$ \csc^2 x = \cot^2 x+ 1 $$
because of Pythagorean Identities, so
$$ \cot^2 x= \csc^2 x-1$$
so
$$ \int \cot^2 x\, \operatorname{d}x= \int \csc^2 x\, \operatorname{d}x - \int 1 \, \operatorname{d}x$$
simplying down to
$$\int \cot^2 x = -\cot x- x + C$$
because I know the derivative of $\cot x$ is $-\csc^2 x$.
I want to know if this is the correct solution because I have many gaps in my logic and frequently make errors and I started learning calculus a couple weeks ago.
If I got something wrong can you please tell me my mistakes.
Thanks, Bot
The result is correct. Don't forget $dx$ in the integrals.