Integral of 1/tan²(x) dx

We want to evaluate the integral

$$ \int \frac{1}{\tan^2(x)} \ dx $$

To tackle this, we can rewrite the integrand in terms of more familiar trigonometric functions.

By the second fundamental identity of trigonometry, the tangent function can be expressed as \( \tan(x) = \frac{\sin(x)}{\cos(x)} \).

Therefore, \( \frac{1}{\tan^2(x)} \) becomes \( \frac{\cos^2(x)}{\sin^2(x)} \).

$$ \int \frac{\cos^2(x)}{\sin^2(x)} \ dx $$

Next, we simplify the numerator using the Pythagorean identity \( \cos^2(x) = 1 - \sin^2(x) \).

Substituting this in, we get:

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

We can now split the expression into two simpler terms:

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

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

The first integral is a standard result: \( \int \frac{1}{\sin^2(x)} \, dx = -\cot(x) \), and the second is simply \( \int 1 \, dx = x \).

So the final result is:

$$ -\cot(x) - x + C $$

where \( C \) is the constant of integration. This gives us the antiderivative of \( \frac{1}{\tan^2(x)} \):

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

And that completes the solution.