In this paper, we will discuss how to do the following indefinite integral problem.

$∫ sin (2x) cos (2x) dx$

Judging from the integrity of sin (2x) cos (2x), we use the substitution method. Since the derivative of sin (2x) is 2cos (2x), then we can assume u = sin (2x):

If u = sin(2x) then du = 2cos(2x) dx or $ \frac{1}{2} du = cos(2x) $ so we get:

∫ sin (2x) cos (2x) dx

= ∫ u (½ du)= ½ ∫ u du

= ½ (½ u²) + C

= $ \frac{1}{4}$ sin² (2x) + C

We can also assume u = cos (2x) because the derivatives of cos (2x) are -2sin (2x):

If u = cos (2x) then du = -2sin (2x) dx or $ - \frac{1}{2} du = sin(2x) $, so we get:

∫ sin(2x) cos(2x) dx

= ∫ u (-½ du)

= -½ ∫ u du

= -½ (-½ u²) + C

= $ \frac{1}{4} $ cos² (2x) + C

In addition to substitution, we can also use the double-angle formula in trigonometry, that is, by using sin (4x) = 2sin (2x) cos (2x), so we get:

∫ sin(2x)cos(2x) dx

= ∫ ½ sin (4x) dx

= ½∫ sin (4x) dx

= $ \frac{1}{8}$ cos (4x) + C.

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