Question

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

Answer 1

We know that

$2 \: sin \: A\: cos \: B = sin(A + B) + sin \: (A - B) \\ \\ so \\ \:\int \sin 3x.\cos 2x \ dx \\ \\ = \int \frac{2}{2} \sin 3x.\cos 2x \ dx \\ \\ \frac{1}{2} \int sin(3x+ 2x) + sin \: (3x - 2x)dx \\ \\ \frac{1}{2} \int sin(5x) + sin \: (x)dx \\ \\ \frac{1}{2} \int (sin(5x) + sin \: ( x))dx$
apply linearity

$\frac{1}{2} \int sin(5x) dx +\frac{1}{2} \int \: sin \: ( x)dx \\ \\ = \frac{ - cos \: 5x}{10} - \frac{cos \: x}{2} + C \\ \\ \int \sin 3x.\cos 2x \ dx=-\frac{cos \:x}{2} - \frac{cos \: 5x}{10} + C$
Hope it helps you.