Question

what is the antiderivative of 2(cos(2x))^2 please with steps!

Answer 1

\[2 \int\limits \cos^2 (2x) dx\]\[u = 2x, du = 2dx\]\[2 \int\limits\limits \cos^2 (2x) dx = \int\limits \cos^2 (u) du\]

Answer 2

Now use a trig. identity.\[\cos (2u) = 2 \cos^2 (u) - 1\]\[\int\limits\limits \cos^2 (u) du = \int\limits \frac {1}{2} \cos (2u) + \frac {1}{2} du\]\[\int\limits\limits \frac {1}{2} \cos (2u) + \frac {1}{2} du = \frac {1}{4} \sin (2u) + \frac {u}{2} + C\]Now just substitute back for u.\[2 \int\limits\limits \cos^2 (2x) dx = \frac {1}{4} \sin (4x) + x + C\]

Answer 3

ok, i dont like that way of doing it, i rather make cos^2x=1/2(1-cos(2x))

Answer 4

but your way is right too, i just wished to confirm, because on wolfram you get a much bigger answer for something more complex but i did it in 2 steps thast why but ya your are right as well, if you want try it with that rule and it also works. thank you though man :)

Answer 5

I would've done this in 1 or 2 steps, but you wanted to see all the steps, so...

Answer 6

\[\cos^2(2x)=\frac{\cos(4x)+1}{2}\]