Question

Find the integrals (primitives):
$\rm \displaystyle\int \sin 4x.\cos 3x\ dx$

Answer 1

We know that

$2 \: sin \: A\: cos \: B = sin(A + B) + sin \: (A - B) \\ \\ so \\ \:\int \sin 4x.\cos 3x \ dx \\ \\ = \int \frac{2}{2} \sin 4x.\cos 3x \ dx \\ \\ \frac{1}{2} \int sin(4x+ 3x) + sin \: (4x - 3x)dx \\ \\ \frac{1}{2} \int sin(7x) + sin \: (x)dx \\ \\ \frac{1}{2} \int (sin(7x) + sin \: ( x))dx$
apply linearity

$\frac{1}{2} \int sin(7x) dx +\frac{1}{2} \int \: sin \: ( x)dx \\ \\ = \frac{ - cos \: 7x}{14} - \frac{cos \: x}{2} + C \\ \\ \int \sin 4x.\cos 3x \ dx=-\frac{cos \:x}{2} - \frac{cos \: 7x}{14} + C$
Hope it helps you.