Question

How do you find `int -arctan(cotx) dx`?

Answer 1

`int(-tan^{-1}(cot)x))dx=x^2/2-\pi/2x+c`

Explanation:

Given we have to find the integral of the tan inverse of cotangent of x, i.e `int(-tan^{-1}(cot(x))dx`

Firstly, we need to simplify the equation.
Remember from inverse trig classes that `tan^{-1}(\theta)+cot^{-1}(\theta)=\pi/2`
So, that means, `tan^{-1}(\theta)=\pi/2-cot^{-1}(\theta)`
So, now we see that we have made our problem a lot easier. So, in the above main equation, if we consider `\theta` as `cot(x)` we got our seemingly very hard problem, become a very easy one.

So, `int(-tan^{-1}(cot(x)))dx=\int(cot^{-1}(cot(x))-pi/2)dx`
(I assume you noticed that I also involved the minus sign that the original question had).
Now, we see that the entire thing becomes easier as the inverse of a function applied to a function is the value itself,
i.e `cot^{-1}(cot(x))=x`
So, out problem turns out to be `int(x-\pi/2)dx`
The answer to this, as you can see, is provided above.