Math Integration question?

Question

Pls help me with this. THANKS!!

1. integral of sin 8x* cos 3x dx

2. integral of sin^3 12x dx

Answer 1

1. use conversion of product to sum... (this formula is not easy to remember though...)
∫ sin 8x cos 3x dx = ½ ∫ [sin 11x + sin 5x] dx
= - 1/22 cos 11x - 1/10 cos 5x + C

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2. ∫ (1 - cos²(12x)) [sin(12x)dx] ... let y = cos(12x) , dy = -12sin(12x)dx

-1/12 ∫(1-y²) dy = -1/12(y - ⅓y³)
= -1/12[cos(12x) - ⅓ cos³(12x)] + C

Answer 2

Question 1
I = ∫ sin 8x cos 3x dx
I = (1/2) ∫ sin(11x) + sin (5x) dx
I = (1/2) [(- 1/11) cos(11x) - (1/5)cos(5x)] + C
I = (-1/22) cos (11x) - (1/10) cos(5x) + C

Question 2
I = ∫ sin ³ (12 x) dx
I = ∫ sin ² (12 x) sin (12 x) dx
I = ∫ (1 - cos ² (12 x)) sin (12 x) dx
Let u = cos 12x
du = - 12 sin 12 x dx
(-1/12) du = sin 12 x dx
I = (-1/12) ∫(1 - u²) du
I = (-1/12) [ u - u³ / 3 ] + C
I = (-1/12) [ cos 12x - (1/3) cos³ 12x ] + C

Answer 3

This is what Maple gives.

1. (-1/22)*cos(11*x) + (1/10)*sin(5*x)

2. (1/36)*(sin(12*x)^2)*cos(12*x) -
(1/18)*cos(12*x) =
(1/36)*(-3+cos(12*x)^2))*cos(12*x)