Find the integrals of the functions in Exercises: 2x 3x - Vidyadip

Question

Find the integrals of the functions in Exercises:
$\sin\text{x } \sin2\text{x }\sin3\text{x}$

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Answer

It is known thtat, $\sin\text{A}\sin\text{B}=\frac{1}{2}\big\{\cos(\text{A}-\text{B})-\cos(\text{A}+\text{B})\big\}$
$\therefore\int\sin\text{x }\sin2\text{x }\sin3\text{x}\text{ dx}$
$=\int\bigg[\sin\text{x}\cdot\frac{1}{2}\big\{\cos(2\text{x}-3\text{x})-\cos(2\text{x}+3\text{x})\big\}\bigg]\text{dx}$
$=\frac{1}{2}\int\big(\sin\text{x}\cos(-\text{x})-\sin\text{x}\cos5\text{x}\big)\text{dx}$
$=\frac{1}{2}\int\big(\sin\text{x}\cos\text{x}-\sin\text{x}\cos5\text{x}\big)\text{dx}$
$=\frac{1}{2}\int\frac{\sin2\text{x}}{2}\text{ dx}-\frac{1}{2}\int\sin\text{x}\cos5\text{x}\text{ dx}$
$=\frac{1}{4}\bigg[\frac{-\cos2\text{x}}{2}\bigg]-\frac{1}{2}\int\bigg\{\frac{1}{2}\sin(\text{x}+5\text{x})+\sin(\text{x}-5\text{x})\bigg\}\text{ dx}$
$=\frac{-\cos2\text{x}}{8}-\frac{1}{4}\int\big(\sin6\text{x}+\sin(-4\text{x})\big)\text{ dx}$
$=\frac{-\cos2\text{x}}{8}-\frac{1}{4}\bigg[\frac{-\cos6\text{x}}{3}+\frac{\cos4\text{x}}{4}\bigg]+\text{C}$
$=\frac{-\cos2\text{x}}{8}-\frac{1}{8}\bigg[\frac{-\cos6\text{x}}{3}+\frac{\cos4\text{x}}{2}\bigg]+\text{C}$
$=\frac{1}{8}\bigg[\frac{\cos6\text{x}}{3}-\frac{\cos4\text{x}}{2}-\cos2\text{x}\bigg]+\text{C}$

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