Wednesday, July 11, 2018

Integral of sin 2x cos 2x

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.

In conclusion, ∫ sin (2x) cos (2x) dx has 3 ways of completion. The answer of these three ways is the equivalent form. Hopefully the writing how to work on the Integral problem of sin 2x cos 2x dx useful for the reader.

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