∫(x * Sin x * Cos x )dx?

Answer 1

To simplify this problem, use the trig identity:

sin (2x) = 2 sin(x)cos(x)

So, your integral becomes:

(1/2)∫(x sin (2x) )dx

Solve using integration by parts and chain rule.

let u = x, dv = sin (2x) dx, so du = dx and v = -(1/2)cos (2x)

∫udv = uv- ∫vdu

(1/2) *{(x(-(1/2)cos (2x)) - ∫-(1/2)cos (2x) dx}

(1/2) *{(x(-(1/2)cos (2x)) + (1/2)∫cos (2x) dx}

(1/2) *{(x(-(1/2)cos (2x)) + (1/2)((1/2) sin (2x))}

(1/2) *{(-1/2) x cos (2x) + (1/4) sin (2x)}

Your answer is:

(1/8) sin (2x) - (1/4) x cos (2x)

Answer 2

Integrate by parts. Let u = x. So du = dx.
Now let dv = sin(x) cos(x) dx.
Use the trig identity sin(2x) = 2sin(x)cos(x).
So dv = (1/2)sin(2x) dx, and v = -(1/4)cos(2x) dx.

uv - ∫ v du =
-(x /4)cos(2x) + ∫ (1/4)cos(2x) dx
-(x /4)cos(2x) + (1/8)sin(2x) + c

Answer 3

∫(x * sin x * cos x )dx
(1/8)∫(2x * sin 2x d2x
= (1/8)[-cos 2x * 2x + ∫-sin 2x d2x
= (1/8)[-cos 2x * 2x + cos 2x]

Answer 4

sinx*cosx=1/2sin2x
1/2Int x*sin2x= 1/2(-1/2x*cos2x+1/2Intcos2x dx)=
=-1/4x*cos2x+1/8 sin2x +C