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anonymous

  • 5 years ago

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

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

  2. anonymous
    • 5 years ago
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    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

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