using double angle formulas for integration

Hi guys
I'm trying to integrate sin^2x dx using cos double angle formula but i keep getting wrong answers that what the text book says can you please see my working and help me if I'm wrong anywhere in it

cos(2x) divided by -2 is not -cos(x), it is (-1/2)cos(2x), you cannot cancel the 2 in the 2x by dividing by 2.

Does that mean you cant write cos(2x)=2cos(x) and 2cos(x)/ -2 will give you -cos(x) ?

Does that mean you cant write cos(2x)=2cos(x) and 2cos(x)/ -2 will give you -cos(x) ?

remember doing transformations of graphs in Core 1 or Core 2 (i cant remember which)

If you have $\displaystyle f(x)$ then $\displaystyle f(ax)$ is a horizontal scalar eg $\displaystyle f(2x)$ is $\displaystyle f(x)$ "squeezed" into half its original width?

and $\displaystyle af(x)$ is a vertical scalar eg $\displaystyle 2f(x)$ "stretches" the graph into 2 times its original height?

so you can clearly say that $\displaystyle f(2x) \neq 2f(x)$ (almost all the time)

so therefore $\displaystyle cos(2x) \neq 2cos(x)$

don't forget to use Radions

Dosent make any difference here does it? As question itself dosent have any radians involves

i think he means in general, definite integrals involving trig are done in radians