Question

What are three different possible equivalent answers for the below? `int`sin(3x)cos(3x)dx

`int`sin(3x)cos(3x)dx

Answer 1

`I = -1/12cos(6x) + C`

Explanation:

Here's how I would do it (1 approach). Letting `u = 3x`, we get `du = 3dx` and `dx =1/3du`.

`I = 1/3int sin(u)cos(u)du`

Now recall that `sin(2alpha) = 2sinalphacosalpha`, thus

`I = 1/3int 1/2sin(2u) du`

`I = 1/6int sin(2u) du`

Now we let `t = 2u`. Thus `dt = 2du` and `du = (dt)/2`.

`I = 1/6int sint (dt)/2`

`I = 1/12 int sint dt`

This is a straightforward integration.

`I = -1/12cost + C`

Now just reverse your substitutions.

`I = -1/12cos(2u) + C`

`I = -1/12cos(6x) + C`

Hopefully this helps!