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If you look in most calculus books there is a table of intergals. Your problem is one of the trigonometric forms. Below is given the reduction formula,
∫sinn(u) du = (-1/n)sinn-1(u)cos(u) + ((n-1)/n)∫sinn-2(u) du
where n ≥ 2 is an integer.
Basically yours would be n=7 & u=3x. You keep integrating until you lose the last integral sign on the right side. Pretty tedious work, but if I am using the right integral form & didn't make any mistakes the answer should be:
(-1/7)sin6(3x)cos(3x) - (6/35)sin4(3x)cos(3x) - (8/35)sin2(3x)cos(3x) - (16/35)cos(3x) + C
Here is the work:
Let u = 3x
u' = 3 dx, so that dx = du/3
∫sin7 3x dx = (1/3)∫sin7(u) du
∫sin7(u) du = (-1/7)sin6(u)cos(u) + (6/7)∫sin5(u) du
= (-1/7)sin6(u)cos(u) + (6/7)[(-1/5)sin4(u)cos(u) + (4/5)∫sin3(u) du]
= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) + (24/35)∫sin3(u) du
= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) + (24/35)[(-1/3)sin2(u)cos(u) + (2/3)∫sin(u) du]
= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) + (16/35)∫sin(u) du
= (-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) - (16/35)cos(u) + C
so that
∫sin7 3x dx = (1/3)∫sin7(u) du
= (1/3 [(-1/7)sin6(u)cos(u) - (6/35)sin4(u)cos(u) - (8/35)sin2(u)cos(u) - (16/35)cos(u)] + C
= -(1/21)sin6(u)cos(u) - (2/35)sin4(u)cos(u) - (8/105)sin2(u)cos(u) - (16/105)cos(u) + C
= -(1/21)sin6(3x)cos(3x) - (2/35)sin4(3x)cos(3x) - (8/105)sin2(3x)cos(3x) - (16/105)cos(3x) + C
Or
∫sin7 3x dx = ∫sin63x sin 3x dx = ∫(sin2 3x)3 sin 3x dx =∫(1 - cos2 3x)3 sin 3x dx
Let u = cos 3x
u' = (cos 3x)'
du = -sin 3x*3 dx, and dx = du/-3sin 3x
= (-1/3)∫(1 - u2)3 du = ∫(1 - 3u2 + 3u4 + u6) du = (-1/3) [u - (3/3)u3 + (3/5)u5 - u7/7] + C
= (-1/3)u + (1/3)u3 - (1/5)u5 + (1/21)u7 + C
= (-1/3)cos 3x + (1/3)cos3 3x - (1/5)cos5 3x + (1/21)cos7 3x + C
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