How do you find arctan(cotx)dx?

1 Answer
Jan 21, 2016

(tan1(cot)x))dx=x22π2x+c

Explanation:

Given we have to find the integral of the tan inverse of cotangent of x, i.e (tan1(cot(x))dx

Firstly, we need to simplify the equation.
Remember from inverse trig classes that tan1(θ)+cot1(θ)=π2
So, that means, tan1(θ)=π2cot1(θ)
So, now we see that we have made our problem a lot easier. So, in the above main equation, if we consider θ as cot(x) we got our seemingly very hard problem, become a very easy one.

So, (tan1(cot(x)))dx=(cot1(cot(x))π2)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 cot1(cot(x))=x
So, out problem turns out to be (xπ2)dx
The answer to this, as you can see, is provided above.