Question

How do you find `int cot^2x*secxdx`?

Answer 1

Notice that `cot^2(x)sec(x) = cot(x)csc(x)` and perform a simple substitution to find that
`intcot^2(x)sec(x)dx = -csc(x) + C`

Explanation:

When an easy substitution does not immediately jump out when looking at an integral, often one can manipulate the integrand into a more convenient form.

In this case:

`intcot^2(x)sec(x)dx = int(cos^2(x))/(sin^2(x))1/cos(x)dx`
`= intcos(x)/sin(x)1/sin(x)dx`
`= intcot(x)csc(x)dx`

We could go directly to the answer from here, as we know that `csc(x)` is the antiderivative of `-csc(x)cot(x)` but let's do the substitution just to see how it works out.

Let `u = csc(x) => du = -csc(x)cot(x)dx`

`=> intcot(x)csc(x)dx = int-du`
`= -intdu`
`= -u+C`
`= -csc(x) + C`

Thus `intcot^2(x)sec(x)dx = -csc(x)+C`