Question

How do I work out `intcot(x)dx` by substitution?

Answer 1

The key to this question is to use your trig identities to rewrite the integral:

`int cot(x) dx = int cos(x) / sin(x) dx`

Now remember that what we are trying to find with substitution is some term being multiplied by it's derivative, which allows us to make use of a variant of the Chain Rule for derivatives that lets us work backwards:

`F'(x) = f'(g(x))g'(x)`

`F(x) = int f'(g(x))g'(x)`

`F(x) = int f'(u)du` where `u = g(x)`

Setting `u = cos(x)` doesn't seem to help us because the derivative of `cos(x)` is `-sin(x)`. Setting `u = sin(x)`, however, will make this one work. It won't always be immediately apparent what we should set `u` to in all integrals solvable by substitution, so sometimes it takes trial and error before you start to remember some patterns.

Setting `u = sin(x)` gives us `du = cos(x)`.

Rewriting our integral:

`int cos(x) / sin(x) dx` => `int (du)/u => int u^(-1)du`, where `u = sin(x)`

We must now remember that the Power Rule doesn't apply when the power is `-1`, and that such integrals actually become `ln(x)`

`int u^(-1)du = ln(u) + C`

Plugging our `u=sin(x)` back in:

`int cot(x) dx = ln( sin(x) ) + C`