Question
Integrate the following integrals:
$\int\sin2\text{x}\sin4\text{x}\sin6\text{x dx}$
$\int\sin2\text{x}\sin4\text{x}\sin6\text{x dx}$
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Answer
$\int\sin2\text{x}\sin4\text{x}\sin6\text{x dx}$
$=\frac{1}{2}\int(2\sin2\text{x}\sin4\text{x})\sin6\text{x dx}$
$=\frac{1}{2}\int\big[\cos(2\text{x}-4\text{x})-\cos(2\text{x}+4\text{x})\big]\sin6\text{x dx}$
$=\frac{1}{2}\int\big[\cos(2\text{x})-\cos(6\text{x})\big]\sin6\text{x dx}$
$=\frac{1}{2}\big[\int\cos(2\text{x})\sin(6\text{x})\text{dx}-\int\cos(6\text{x})\sin(6\text{x})\text{dx}\big]$
$=\frac{1}{4}\big[\int2\cos(2\text{x})\sin(6\text{x})\text{dx}-\int2\cos(6\text{x})\sin(6\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big\{\int\big[\sin(2\text{x}+6\text{x})-\sin(2\text{x}-6\text{x})\big]\text{dx}-\int\sin(12\text{x})\text{dx}\Big\}$
$=\frac{1}{4}\Big[\int\sin(8\text{x})\text{dx}+\int\sin(4\text{x})\text{dx}-\int\sin(12\text{x})\text{dx}\Big]$
$=\frac{1}{4}\Big[\frac{-\cos(8\text{x})}{8}+\frac{-\cos(4\text{x})}{4}+\frac{\cos(12\text{x})}{12}\Big]+\text{C}$
$=-\frac{\cos(8\text{x})}{32}-\frac{\cos(4\text{x})}{16}+\frac{\cos(12\text{x})}{48}+\text{C}$
$=\frac{1}{2}\int(2\sin2\text{x}\sin4\text{x})\sin6\text{x dx}$
$=\frac{1}{2}\int\big[\cos(2\text{x}-4\text{x})-\cos(2\text{x}+4\text{x})\big]\sin6\text{x dx}$
$=\frac{1}{2}\int\big[\cos(2\text{x})-\cos(6\text{x})\big]\sin6\text{x dx}$
$=\frac{1}{2}\big[\int\cos(2\text{x})\sin(6\text{x})\text{dx}-\int\cos(6\text{x})\sin(6\text{x})\text{dx}\big]$
$=\frac{1}{4}\big[\int2\cos(2\text{x})\sin(6\text{x})\text{dx}-\int2\cos(6\text{x})\sin(6\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big\{\int\big[\sin(2\text{x}+6\text{x})-\sin(2\text{x}-6\text{x})\big]\text{dx}-\int\sin(12\text{x})\text{dx}\Big\}$
$=\frac{1}{4}\Big[\int\sin(8\text{x})\text{dx}+\int\sin(4\text{x})\text{dx}-\int\sin(12\text{x})\text{dx}\Big]$
$=\frac{1}{4}\Big[\frac{-\cos(8\text{x})}{8}+\frac{-\cos(4\text{x})}{4}+\frac{\cos(12\text{x})}{12}\Big]+\text{C}$
$=-\frac{\cos(8\text{x})}{32}-\frac{\cos(4\text{x})}{16}+\frac{\cos(12\text{x})}{48}+\text{C}$
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