Question

Find the integrals (primitives):
$\rm \displaystyle\int \sin x.\sin 2x.\sin 3x \ dx$

Answer 1

We know that

$\sin(3x) = 3sin \: x - 4 {sin}^{3} x \\ \\ \sin(2x) = 2sin \: x \: cos \: x$
$\int \sin x.\sin 2x.\sin 3x \ dx \\ \\ = \int \sin x.2 \: sin \: x \: cos \: x.(3 \: sin \: x - 4 {sin}^{3} x) \ dx \\ \\ = \int( 6 {sin}^{3} x - 8 {sin}^{5} x)cos \: x \: dx \\ \\ let \\ \\ sin \: x = t \\ \\ cos \: x \: dx = dt \\ \\ = \int( 6 {t}^{3} - 8 {t}^{5} )dt \\ \\ apply \: linearity \\ \\ \int \: 6 {t}^{3} \: dt - \int8{t}^{5}dt \\ \\ = \frac{6 {t}^{4} }{4} - \frac{8 {t}^{6} }{6} + C \\ \\ = \frac{3{t}^{4} }{2} - \frac{4 {t}^{6} }{3} + C\\ \\ redo \: substitution \\ \\ \int \sin x.\sin 2x.\sin 3x \ dx= \frac{3{sin}^{4}x }{2} - \frac{4 {sin}^{6}x }{3} + C$
Hope it helps you.