What is the integral of #int sin^3 3x cos 3x dx#?

1 Answer
Jan 23, 2016

#intsin^3(3x)cos(3x)dx=frac{sin^4(3x)}{12}+c#

Explanation:

Take the given equation #intsin^3(3x)cos3xdx#
Take the function #sin(3x)=t#
Now differentiate with respect to #t# on both sides of this
#\frac{d}{dt}(sin3x)=1\implies3cos3x*\frac{dx}{dt}=1\impliescos3xdx=dt/3#

Substituting the given above values for the main equation, we get
#intt^3/3dt#
This looks easy. Let's directly integrate.
#intt^3/3dt=t^4/(3*4)+c=t^4/12+c#

That's what you want, right? Oh, substitution? That's there up in the answer section.