Question

What is the integral of `int sin(3x+4)dx`?

Answer 1

`-1/3cos(3x+4)+C`

Explanation:

We will want to use the following rule:

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

So, when integrating

`intsin(3x+4)dx`

We let:

`u=3x+4" "=>" "(du)/dx=3" "=>" "du=3dx`

Notice that in the integral, we only have a `dx` term, but we want a `3dx` term. To achieve this, multiply the interior on the integral by `3`. Balance this by multiplying the exterior of the integral by `1/3`.

`intsin(3x+4)dx=1/3intsin(3x+4)*3dx`

Substituting in what we know, which are `3x+4=u` and `3dx=du`.

`1/3intsin(3x+4)*3dx=1/3intsin(u)du`

Using the initial rule, this becomes

`1/3intsin(u)du=-1/3cos(u)+C=-1/3cos(3x+4)+C`