Question

Integral of 2sin(2x)cos(2x) dx =
Ok... so I used the fact that sin 2x = 2sinxcosx to treat 2x here as x, and so I simplified it to sin 4x
If I did everything right. But I am not too sure what to do next... Or what to do with integral of trigs that have functions inside of them. Can anyone point me the way?

Answer 1

This just looks like a simple u-substitution

Answer 2

\[\large 2\int^{}_{} sin(2x)cos(2x)dx\]

Answer 3

Let

u = 2x
du = 2dx
dx = du
---
2

Answer 4

\[\large \frac{2}{2}\int^{}_{} sin(u)cos(u)du\]

Answer 5

\[\large \int^{}_{} sin(u)cos(u)du\]

Answer 6

Now make another substitution....

s = sin(u)
ds = cos(u)du

\[\large \int^{}_{} s \space ds\]

Answer 7

Integral of s is

\[\large \frac{s^2}{2} + C\]

Answer 8

Now re-substitute in the original values...

s = sin(u)

\[\large \frac{sin(u)^2}{2} + C\]

Answer 9

And u = 2x so

\[\large \frac{1}{2}\sin^2(2x) + C\]

Answer 10

Which then simplifies down to

\[\large \frac{1}{4}sin(4x) + C\]

Answer 11

Wait. where did the 1/2 come from after you plug in u?

Answer 12

It doesn't have the du for it so... why?

Answer 13

\(\huge \frac{sin^2(u)}{2} = \frac{1}{2}sin^2(u)\)

Just decided to move the 1/2 out in front instead of dividing the whole thing by 2

Answer 14

Ah... Ok, that makes perfect sense then, sorry. Guess I gotta look up the sin^2(2x) identity later... Would my method of sin(4x) have worked?? Because finding the integral of it would be... u method again? u = 4x
uh... would it have?

Answer 15

Indeed ...actually looking back.you would have received the right answer that way too...

\[\large \int^{}_{}sin(4x)dx = -\frac{1}{4}cos(u) + C= -\frac{1}{4}cos(4x) + C\]

same difference there

Answer 16

wouldn't that be wrong though? The answer from sin 4x is cos where as the answer you derived using S is sin... what's going on!?

Answer 17

still best response, thanks.