Question

How do I find the integral of sin^4 (3x)?

Answer 1

Perhaps the following will help:
\[\sin(3x)^4=1/8(3-4 \cos(6x)+\cos(12x))\]

Answer 2

You must simplify \[\sin^{4}(3x)\] using the following half-angle formulas
\[\sin^2(u) = 1/2(1-\cos(2u))\]\[\cos^{2}(u)=1/2(1+\cos(2u))\]
To do this, perform the following steps:
\[\sin^{4}(3x) = \sin^2(3x)\sin^2(3x)\]
\[\sin^2(3x)\sin^2(3x) = 1/2(1-\cos(6x))1/2(1-\cos(6x))\]
\[sin^2(3x)\sin^2(3x) = 1/2[(1-\cos(6x))(1-\cos(6x))]\]
Now FOIL \[(1-\cos(6x))(1-\cos(6x)) = 1-2\cos(6x)+\cos^{2}(6x)\]
\[\cos^{2}(6x) = 1/2(1+\cos(12x))\]
So,
\[\sin^2(3x)\sin^2(3x) = 1/2[1-2\cos(6x)+1/2+1/2\cos(12x)]\]
Now that we have what sin^4(3x) is simplified, we can integrate across the terms with respect to x.
\[1/2[\int\limits_{}^{}1-\int\limits_{}^{}2\cos(6x)+\int\limits_{}^{}1/2+1/2\int\limits_{}^{}\cos12x]\]
After integration:
\[1/2[x-2/6\sin(6x)+1/2x+1/24\sin(12x)]+c\]
I'll leave you to simplify