What is the integral of ##cos2(theta)##?

##int cos(2theta) "d"theta = 1/2 sin(2theta) + C##,

where ##C## is an integration constant.

I think you mean ##cos(2theta)## instead of ##cos2(theta)##.

If you know that ##int cos(x) dx = sin(x) + C##, then we can use a (which is the reverse of the ).

Let ##u = 2theta##,

##frac{"d"u}{"d"theta} = 2##.

So,

##int cos(2theta) "d"theta = 1/2 int cos(2theta) * (2) "d"theta##

##= 1/2 int cos(2theta) * frac{"d"u}{"d"theta} "d"theta##

##= 1/2 int cos(u) "d"u##

##= 1/2 (sin(u) + C_1)##, where ##C_1## is an integration constant.

##= 1/2 sin(2theta) + C_2##, where ##C_2 = 1/2 C_1##.