Integral Calculation Exercise 26

We want to evaluate the integral

$$ \int \cot (ax+b) \ dx $$

We’ll use substitution, introducing the auxiliary variable u = ax + b:

$$ u = ax + b $$

Taking the differential:

$$ du = a \ dx $$

Solving for $ dx $:

$$ dx = \frac{1}{a} \ du $$

Substituting dx = $\frac{1}{a} \, du$ into the integral gives:

$$ \int \cot (ax + b) \cdot \frac{1}{a} \ du $$

Now, replacing $ ax + b $ with $ u $:

$$ \int \cot (u) \cdot \frac{1}{a} \ du $$

Since $ \frac{1}{a} $ is a constant, we can factor it out:

$$ \frac{1}{a} \int \cot (u) \ du $$

Recall that $ \cot(u) = \frac{\cos(u)}{\sin(u)} $, so the integral becomes:

$$ \frac{1}{a} \int \frac{\cos(u)}{\sin(u)} \ du $$

We now perform a second substitution by letting t = $\sin(u)$:

$$ t = \sin(u) $$

Then:

$$ dt = \cos(u) \ du $$

Solving for $ du $:

$$ du = \frac{1}{\cos(u)} \ dt $$

Substituting into the integral:

$$ \frac{1}{a} \int \frac{\cos(u)}{\sin(u)} \cdot \frac{1}{\cos(u)} \ dt $$

$$ \frac{1}{a} \int \frac{1}{\sin(u)} \ dt $$

Now replacing $ \sin(u) $ with $ t $:

$$ \frac{1}{a} \int \frac{1}{t} \ dt $$

This is a basic integral: $ \int \frac{1}{t} dt = \log |t| + C $

So we have:

$$ \frac{1}{a} \log |t| + C $$

Since $ t = \sin(u) $, we substitute back:

$$ \frac{1}{a} \log |\sin(u)| + C $$

And finally, replacing $ u = ax + b $:

$$ \frac{1}{a} \log |\sin(ax + b)| + C $$

This is the final expression for the antiderivative.

And that completes the solution.