Urgent !!!!! Integrate x^2 sin^2 x cos x By part?

∫ x^2 sin^2(x) cosx dx
=∫ 1/2 x^2 sin2x sinx dx
=1/2 (-1/2) ∫ x^2 (cos3x - cosx) dx
=-1/4 ∫ x^2 cos3x dx +1/4 ∫ x^2 cosx dx

Find ∫ x^2 cos3x dx by part
∫ x^2 cos3x dx = 1/3 sin3x x^2 - 2/3 ∫ xsin3x dx
= 1/3 sin3x x^2 - 2/3 [ -xcos3x / 3 + ∫ cos3x/3 dx]
=1/3 sin3x x^2 +2/9 x cos3x -2/27 sin3x +C

Find ∫ x^2 cosx dx by part
∫ x^2 cosx dx = sinx x^2 - 2∫ x sinx dx
= sinx x^2 -2[-xcosx - ∫ -cosx dx]
=x^2 sinx + 2xcosx - 2sinx + C

so ∫ x^2 sin^2(x) cosx dx = -1/4 [ 1/3 sin3x x^2 +2/9 x cos3x -2/27 sin3x ] +1/4 [x^2 sinx + 2xcosx - 2sinx ]
= -1/12 sin3x x^2 -1/18 x cos3x + 1/54 sin3x +1/4 x^2 sinx + 1/2 x cosx - 1/2 sinx +C

Please tell me if there is a faster method

sin^2 (x) * cos^2 (x) is the same as (sin (x) * cos (x))^2. s
in (2x) = 2 sin (x) * cos (x). Dividing both sides by 2, we find that sin (2x) / 2 = sin (x) * cos (x).
sin^2 (x) * cos^2 (x) = (sin (x) * cos (x))^2 = (sin (2x) / 2)^2 = sin^2 (2x) / 4.
sin^2 (2x) / 4
So sin^2 (2x) = (1 - cos (4x)) / 2.

So sin^2 (2x) / 4 = (1 - cos (4x)) / 8 = 1/8 - 1/8 cos (4x).

Now that you have rewritten the integrand as 1/8 - 1/8 cos (4x), you should be able to integrate it. (You should get x/8 - 1/32 sin (4x) + C.)