How do I integrate cos 2 x?no_redirect=1

i was able to do the 1st question, and the other i can't do the step by step solution.

int. x(cosx)^2 dx = int. X(1/2cos2X + 1/2) dx ---> (1)

= int. 1/2Xcos2X dx + int. 1/2X dx -->(2)

= (Xsin2X)/4 - int. (sin2X/4) dx + int. (1/2x) dx

= (Xsin2X)/4 - 1/4 int. (sin2X) dx + int. (1/2X) dx

= 1/4 (Xsin2X) - (1/4)(-1/2 cos2X) + (1/4 X^2) + C

The answer is = 1/4(Xsin2X) - 1/8(cos2X) + 1/4X^2 + C, where C is a

constant.

Elaboration:

(1) You have to apply the double angle formula:

cos2X = 2(cosX)^2 - 1

cos2X + 1 = 2(cosX)^2

(cosX)^2 = (1/2cos2X) - 1/2

The reason you have to apply this formula is that you can't directly integrate (cosX)^2. You can only integrate cos2X.

(2) You have to integrate the expression by using the By-Parts method.

i.e.

int. 1/2X(cos2X) dx

When using the By-Parts method, you have to

Let u = 1/2X >>> du/dx = 1/2

Let dv/dx = cos2X >>> v = 1/2sin2X + C

Hence, the formula for By-Parts is: uv - int. v(du/dx)

int. 1/2X(cos2X) dx

= (1/2x)(1/2sin2X) - int. 1/2(1/2sin2X)

= (1/4sin2X) - int. (1/4sin2X)

I hope this may help you ^_^