Find the integral of sin u sin 2 u sin 3 u .

Question

Find the integral of $\sin u \sin 2 u \sin 3 u$ .

Smart Achievers · Accepted Answer

(3)

Ans. $\frac{1}{8} [\frac{\cos 6 u}{3} - \frac{\cos 4 u}{2} - \cos 2 u] + C$

Explanation:

The given integral is,

$\int \sin u \sin 2 u \sin 3 u d u .$

We know that,

$\sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)]$

Using the above formula for $\sin 2 u \sin 3 u$ in the given integral, we get

$\int \sin u \sin 2 u \sin 3 u d u$

$= \int \sin u \cdot \frac{1}{2} [\cos (2 u - 3 u) - \cos (2 u + 3 u)] d u$

$= \frac{1}{2} \int [\sin u \cos (- u) - \sin u \cos 5 u] d u$

$= \frac{1}{2} \int (\sin u \cos u - \sin u \cos 5 u) d u$

We know that,

$2 \sin A \cos A = \sin 2 A$

and $\sin A \cos B = \frac{1}{2} [\sin (A + B) + \sin (A - B)]$

Using the above two formulas in the integral, we get

$\frac{1}{2} \int (\sin u \cos u - \sin u \cos 5 u) d u$

$= \frac{1}{2} \int \frac{2 \sin u \cos u}{2} d u - \frac{1}{2} \int \frac{1}{2} [\sin (u + 5 u) + \sin (u - 5 u)] d u$

$= \frac{1}{4} \int \sin 2 u d u - \frac{1}{4} \int [\sin 6 u + \sin (- 4 u)] d u$

$= \frac{1}{4} [\int \sin 2 u d u - {\int \sin 6 u d u - \int \sin 4 u d u}]$

$= \frac{1}{4} [\frac{- \cos 2 u}{2} - (\frac{- \cos 6 u}{6} + \frac{\cos 4 u}{4})] + C$

$= \frac{1}{4} \times \frac{1}{2} [- \cos 2 u + \frac{\cos 6 u}{3} - \frac{\cos 4 u}{2}] + C$

$= \frac{1}{8} [\frac{\cos 6 u}{3} - \frac{\cos 4 u}{2} - \cos 2 u] + C$

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