Question

Find the integrals (primitives):
$\rm \displaystyle\int \frac{\sin 4x}{\cos 2x}\ dx$

Answer 1

We know that

$sin \: 2x = 2 \: sin \: x \: cos \: x$

Then

$\frac{sin \: 4x}{cos \: 2x} = \frac{2 \: sin \: 2x \: cos \: 2x}{cos \: 2x} \\ \\ = 2sin \: 2x$
So,
$\int \frac{\sin 4x}{\cos 2x}\ dx =\int 2 \: sin \: 2x \: dx \\ \\ =2\int \: sin \: 2x \: dx$
$let \: 2x = t \\ \\ 2dx = dt \\ \\ dx = \frac{1}{2} dt$
Substitute the value

$= 2 \times \frac{1}{2} \int \: sin \: t \: dt \\ \\ = \int \: sin \: t \: dt \\ \\ = - cos \: t + C \\ \\ undo \: substitution \\ \\ \int \frac{\sin 4x}{\cos 2x}\ dx = - cos \: 2x \: + C$
Hope it helps you.