Here you will learn proof of integration of cot inverse x or arccot x and examples based on it.
Let’s begin –
Integration of Cot Inverse x
The integration of cot inverse x or arccot x is \(xcot^{-1}x\) + \(1\over 2\) \(log |1 + x^2|\) + C
Where C is the integration constant.
i.e. \(\int\) \(cot^{-1}x\) = \(xcot^{-1}x\) – \(1\over 2\) \(log |1 + x^2|\) + C
Proof :
We have, I = \(\int\) \(cot^{-1}x\) dx
Let \(cot^{-1}x\) = t,
Then, x = cot t
\(\implies\) dx = d(cot t) = \(-cosec^2 t\) dt
\(\therefore\) I = \(\int\) \(cot^{-1}x\) dx
\(\implies\) I = \(\int\) t \(-cosec^2 t\) dt
By using integration by parts formula,
I = t cot t – \(\int\) 1. (cot t) dt
I = t cot t – log |sin t| + C
Since cot t = x \(\implies\) cosec t = \(\sqrt{1 + cot^2 t}\) = \(\sqrt{1 + x^2}\)
Hence, sin t = \(1\over \sqrt{1 + x^2}\)
Now, Put t = \(cot^{-1}x\) here,
\(\implies\) I = x \(cot^{-1}x\) – \(log |{1\over \sqrt{ 1+ x^2}}|\) + C
Hence, \(\int\) \(cot^{-1}x\) dx = \(xcot^{-1}x\) + \(1\over 2\) \(log |1 + x^2|\) + C
Example : Evaluate \(\int\) \(x cot^{-1} x\) dx
Solution : We have,
I = \(\int\) \(x cot^{-1} x\) dx
By using integration by parts formula,
I = \(cot^{-1} x\) \(x^2\over 2\) – \(\int\) \(-1\over 1 + x^2\) \(\times\) \(x^2\over 2\) dx
I = \(tan^{-1} x\) \(x^2\over 2\) + \(1\over 2\) \(\int\) \(x^2 + 1 – 1\over 1 + x^2\)dx
= \(x^2\over 2\) \(cot^{-1} x\) + \(1\over 2\) \(\int\) 1 – \(1\over 1 + x^2\)dx
\(\implies\) I = \(x^2\over 2\) \(cot^{-1} x\) + \(1\over 2\) (x – \(tan^{-1} x\)) + C
\(\implies\) I = \(x^2\over 2\) \(cot^{-1} x\) + \(x\over 2\) – \(tan^{-1} x\over 2\) + C
Related Questions
What is the Differentiation of cot inverse x ?
