Integration of cot inverse x

Integration of cot inverse x can be calculated using integration by parts .Here is the formula for it

\[ \int \cot^{-1}x \, dx = x \cot^{-1}x + \frac {1}{2} \ln (1 + x^2) + C \]

where (C) is the constant of integration.

Proof of integration of cot inverse x

To find the integral we use integration by parts. Integration by parts is based on the product rule for differentiation and is given by:

$\int f(x) g(x) dx = f(x) (\int g(x) dx )- \int \left \{ \frac {df(x)}{dx} \int g(x) dx \right \} dx $
In our case, we can let $ f(x) = \cot^{-1}x $ and $ g(x) = 1 $. Then

• $ \frac {df(x)}{dx} = -\frac{1}{1+x^2} \, dx $
• $\int g(x) dx = \int dx = x $

Now, substitute these into the integration by parts formula:

$$
\int \cot^{-1}x \, dx = x \cot^{-1}x – \int x \cdot \frac{-1}{1+x^2} \, dx
$$

$ =x \tan^{-1}x + \int \frac{x}{1+x^2} \, dx \\
=x \tan^{-1}x – \frac {1}{2} \int \frac{2x}{1+x^2} \, dx $

Now lets calculate the second integral separately $\int \frac{2x}{1+x^2} \, dx $

Let $t= 1+x^2$
then $dt=2x dx$
Therefore

$\int \frac{2x}{1+x^2} \, dx = \int \frac{1}{t} \, dx \\
= \ln |t| = \ln |1+x^2|$

Substituting this value in main integral , we get

$$
\int \cot^{-1}x \, dx = x \cot^{-1}x + \frac {1}{2} \ln (1 + x^2) + C $$

Definite Integral of tan inverse x

To find the definite integral of $\cot^{-1}x$ over a specific interval, we use the same approach as with the indefinite integral, but we’ll apply the limits of integration at the end.

The definite integral of $\cot^{-1}x$ from $a$ to $b$ is given by:

$$\int_{a}^{b} \cot^{-1}x \, dx = b \cot^{-1}b – a \cot^{-1}a + \frac {1}{2}\ln (1 + b^2) – \frac {1}{2} \ln (1 + a^2) $$

This expression represents the accumulated area under the curve of $\cot^{-1}x$ from $x = a$ to $x = b$.

Solved Examples on Integration of tan inverse x

Question 1

$$\int_{0}^{1} \cot^{-1}x \, dx$$

Solution

$\int_{0}^{1} \cot^{-1}x \, dx = [x \cot^{-1}x + \frac {1}{2} \ln (1 + x^2)] _0 ^1 \\
= 1. cot^{-1} 1+ \frac {\ln 2}{2} – 0 – \frac {\ln 1}{2} = \frac {\pi}{4} + \ln 2/2$

Question 2

$$ \int x cot^{-1} x \, dx$$

Solution

Applying integration by parts
$\int u \, dv = uv – \int v \, du$.
Let $u = \cot^{-1}x$. Then, $du = \frac{-dx}{1 + x^2}$.
Let $dv = x \, dx$. Then, $v = \frac{x^2}{2}$.
Then
$ \int x \cot^{-1}x \, dx = \frac{x^2}{2} \cot^{-1}x + \int \frac{x^2}{2} \cdot \frac{dx}{1 + x^2} \
=\frac{x^2}{2} \cot^{-1}x + \frac {1}{2} \int \frac{x^2}{1 + x^2} \; dx$
Now
$\int \frac{x^2}{1 + x^2} \; dx = \int \frac{1+x^2 – 1}{1 + x^2} \; dx \\
= \int 1 dx – \int \frac {1}{1+x^2} dx \\
= x – tan^{-}x$

Applying these steps to the integral of $x \tan^{-1}x$, we obtained:

$$ \frac{x^2 \cot^{-1}x}{2} + \frac{x }{2} – \frac{\tan^{-1}x}{2} + C $$

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