Question

Find the integrals (primitives):
$\rm \int \cos 2x.\cos 4x\ dx$

Answer 1

here is the answer ✍️✍️✍️✍️✍️

Answer 2

As we know that

$2 \: cos \: A \: cos \: B = cos(A + B) + cos(A - B)$
$\cos 2x.\cos 4x = > \frac{2\cos 2x.\cos 4x}{2} \\ \\ = \frac{1}{2} (cos(2x + 4x) + cos(2x - 4x)) \\ \\ = \frac{1}{2} (cos \: 6x + cos \: 2x) \\ \\ since \: cos( - x) = cos \: x$
Now integrate

$\int\frac{1}{2} (cos \: 6x + cos \: 2x) dx\\ \\ \\ = \frac{1}{2} \int \: cos \: 6x \: dx + \frac{1}{2} \int \: cos \: 2x \: dx \\ \\ = \frac{1}{2} \frac{sin \: 6x}{6} + \frac{1}{2} \frac{sin \: 2x}{2} + C \\ \\ \int \cos 2x.\cos 4x\ dx = \frac{1}{12} sin \: 6x + \frac{1}{4} sin \: 2x + C$
Hope it helps you.