The given integral is:

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

Solution:

We use the identity:

$\cot^2 x = \csc^2 x - 1$

Thus, rewriting the integral:

$\int \cot^2 x \,dx = \int (\csc^2 x - 1) \,dx$

Now, integrating each term separately:

$\int \csc^2 x \,dx - \int 1 \,dx$

We know:

$\int \csc^2 x \,dx = -\cot x$

$\int 1 \,dx = x$

So, the final result is:

$-\cot x - x + C$

where $C$ is the constant of integration.

Would you like a detailed explanation of any step?

Related Questions:

1. How do you derive the identity $\cot^2 x = \csc^2 x - 1$?
2. What is the integral of $\csc^2 x$ and why does it result in $-\cot x$?
3. How can we solve $\int \tan^2 x \,dx$ using a similar approach?
4. What are the applications of the integral $\int \cot^2 x \,dx$ in trigonometry and calculus?
5. How do we evaluate the definite integral $\int_a^b \cot^2 x \,dx$ over specific limits?

Tip:

When integrating trigonometric functions, look for trigonometric identities that simplify the expression before integrating directly!