## anonymous 5 years ago How do I find the integral of sin^4 (3x)?

1. anonymous

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

2. anonymous

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