Question

What is the integral of `int sin(2x)dx`?

Answer 1

`-1/2cos(2x)+C`

Explanation:

We will use the rule:

`intsin(u)du=-cos(u)+C`

So, if we have

`intsin(2x)dx`

we should set

`u=2x" "=>" "du=2dx`

In order to have a `2dx` in the integral expression, multiply by a `2` on the inside and balance it with a `1/2` on the outside:

`=1/2intsin(2x)*2dx`

Substitute:

`=1/2intsin(u)du`

Using the initial rule, this becomes:

`=-1/2cos(u)+C`

Since `u=2x`,

`=-1/2cos(2x)+C`