Question

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

Answer 1

`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.