integration of cot square x

The integration of cot square x $\cot ^2 x$, can be found using trigonometry identities and fundamental integral formulas . The integral of $\cot^2 x$ with respect to (x) is:

\[
\int \cot^2 x \, dx =-\cot x -x + C
\]

Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant.

Proof of integration of cot square x

We know from trigonometry identities
$cosec^2 x =1 + \cot^2 x$
or $\cot^2 x =\csc^2 x -1$
$\int \cot^2 x dx = \int (\csc^2 x -1) dx$
Now we know that from fundamental integration formula
$\int ( \csc^2 x) \; dx = -\cot x + C$
Therefore
$$\int \cot^2 x \; dx = \int (\csc^2 x -1) dx= -\cot x -x + C$

Definite Integral of cot square x

To find the definite integral of $\cot^2x$ 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 $\cot^2x$ from $a$ to $b$ is given by:

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

This expression represents the accumulated area under the curve of $\cot^2x$ from $x = a$ to $x = b$.

Solved Examples

Question 1

$$ \int x \cot^2 x \; dx $$

Solution

Using Integration by Parts,

• $u = x$, which implies $du = dx$, and
• $dv = \cot^2(x) \, dx$.

Now we already derive earliar , integrating $dv = \cot^2(x) \, dx$ yields $v = -x – \cot(x)$

With $u = x$ and $v = -x – \cot(x)$, integration by parts gives us:

$$
\int x \cot^2(x) \, dx = uv – \int v \, du
$$

Substituting the values of $u$ and $v$ into this formula:

• $uv = x(-x – \cot(x))$
• To find $\int v \, du$, we integrate $-x – \cot(x)$ with respect to $x$.

The integration by parts formula leads to the solution:

$$
-\frac{x^2}{2} – \frac{x}{\tan(x)} + \log(\sin(x)) + C
$$

Question 2

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

Solution

we can use a trigonometric identity to simplify the expression. The key identity to use here is

$$
1 + \cot^2(x) = \csc^2(x)
$$

which means the integral simplifies to

$$
\int \frac{1}{\csc^2(x)} \, dx = \int \sin^2(x) \, dx
$$

To solve (\int \sin^2(x) \, dx), we can use the half-angle identity:

$$
\sin^2(x) = \frac{1 – \cos(2x)}{2}
$$

So, we get
$$
\int \frac{1}{1 + \cot^2(x)} \, dx = \frac{x}{2} – \frac{\sin(2x)}{4} + C
$$