Integrate the following integrals: 2x 3x dx - Vidyadip

Question

Integrate the following integrals:
$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$

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Answer

$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$
$=\frac{1}{2}\int(2\sin\text{x}\cos2\text{x})\sin3\text{x dx}$
$=\frac{1}{2}\int\big[\sin(\text{x}+2\text{x})+\sin(\text{x}-2\text{x})\big]\sin(3\text{x) dx}$
$=\frac{1}{2}\int\big[\sin(3\text{x})-\sin(\text{x})\big]\sin(3\text{x) dx}$
$=\frac{1}{2}\big[\int\sin^2(3\text{x})\text{dx}-\int\sin(\text{x})\sin(3\text{x})\text{dx}\big]$
$=\frac{1}{4}\big[\int2\sin^2(3\text{x})\text{dx}-\int2\sin(\text{x})\sin(3\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big\{\int\big[1-\cos(6\text{x})\big]\text{dx}-\int\big[\cos(\text{x}-3\text{x})-\cos(\text{x}+3\text{x})\big]\text{dx}\Big\}$
$=\frac{1}{4}\big[\int1\text{dx}-\int\cos(6\text{x})\text{dx}-\int\cos(2\text{x})\text{dx}+\int\cos(4\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big[\text{x}-\frac{\sin(6\text{x})}{6}-\frac{\sin(2\text{x})}{2}+\frac{\sin(4\text{x})}{4}\Big]+\text{C}$
$=\frac{\text{x}}{4}-\frac{\sin(6\text{x})}{24}-\frac{\sin(2\text{x})}{8}+\frac{\sin(4\text{x})}{16}+\text{C}$
Hence, $\int\sin\text{x}\cos2\text{x}\sin3\text{x}\text{ dx}$ $=\frac{\text{x}}{4}-\frac{\sin(6\text{x})}{24}-\frac{\sin(2\text{x})}{8}+\frac{\sin(4\text{x})}{16}+\text{C}$

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