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

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

1 Answer
Feb 21, 2018

#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!