# Integration of Cot Inverse x

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 ?

What is the Integration of Cotx ?

What is the integration of sec inverse root x ?