Find the integral of 3cos2x-2sinx

Answer 1

If your question means:

∫ [ 3 cos ( 2 x ) - 2 sin x ] dx

then

∫ cos ( 2 x ) dx

Substitute:

u = 2 x

du = 2 dx

du / 2 = dx

dx = du / 2

∫ cos ( 2 x ) dx = ∫ cos u ∙ du / 2 = 1 / 2 ∫ cos u ∙ du = 1 / 2 sin u = 1 / 2 sin ( 2 x )

∫ sin x dx = - cos x

∫ [ 3 cos ( 2 x ) - 2 sin x ] dx =

3 ∫ cos ( 2 x ) dx - 2 ∫ sin x dx = 3 / 2 sin ( 2 x ) - 2 ( - cos x ) + C =

3 / 2 sin ( 2 x ) + 2 cos x + C

Answer 2

Recall that cos (2x) = 2 cos^2 x - 1
or
cos^2 x = (cos 2x + 1)/2 = (1/2)cos 2x + 1/2

∫ 3cos2x-2sinx dx
= ∫ (3/2)cos 2x + 3/2 - 2sinx ) dx
= 3/4 sin 2x + (3/2)x + 2cosx + c

Answer 3

Just goes to show you how important brackets are when typing
in the required format on this webpage.

Bosnian interpreted your 3cos2x
as 3cos(2x), and his answer is correct according to that

I interpreted your 3cos2x
as 3cos^2 (x), and my answer reflects that way of looking at it.

Make sure that you make it clear which way.

Answer 4

Tnx