Conflicting integrations of cos²x?

Question

Method 1:
Now cos²x = (cosx)², and integrating normally I get (-cos³x)/(3sinx).

Method 2:
Using identities:
cos²x + sin²x = 1 and
cos²x - sin²x = cos(2x) we get
2cos²x = 1 + cos (2x) and
cos²x = 1/2 (1+cos(2x))
which upon integrating gives x/2 - (sin(2x))/4.

These two answers are clearly different, but why?
I think I must have made some silly error the first way, but I don't know where.

Answer 1

the first one is wrong....
actually we integrate a function of x WITH RESPECT TO x
that is,
if you consider cos x as some t,so that your function is t squared,
we also account for dx,that is
since t=cos x
dt/dx = -sin x
dx=dt/(-sin x)
=dt/(-Sqroot(1-cos^2 x)
=dt/(-sqroot(1-t^2))
so the integral is
integration of(-t^2/(sqroot(1-t^2))) with respect to t
...
the second method is easier

Answer 2

As others have said, your first method is wrong. The second method is OK but you made a mistake, the right answer is x/2 + (sin(2x)/4 + C , where C is a constant. And note you changed the sign of (sin(2x)/4

Answer 3

The 1st method is integration with respect to cos x and not x