What is the integral of #int sin(2x)dx#?
1 Answer
Mar 17, 2016
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
#=1/2intsin(2x)*2dx#
Substitute:
#=1/2intsin(u)du#
Using the initial rule, this becomes:
#=-1/2cos(u)+C#
Since
#=-1/2cos(2x)+C#