Trig Integral (sin/cos)?
Indefinite integral of sin(18x)^3 * cos(18x)^5
I let u = cos x and du = -sin x and got:
18 int ((1-cos^2(x)*sin(x)*cos^2(x)*cos^2(x)*cos(x) dx
= -18 int (1-u^2) u^5 du = -18 int (u^5 - u^7) du = -18 ( u^6/6 - u^8/8)
= -18(cos^6(x)/6 - cos^8(x)/8)
But, the -18 is incorrect. It has something to do with my initial breakdown of the problem. What did I miss?
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∫ sin^3(18x) cos^5(18x) dx
= ∫ sin^3(18x) cos^4(18x) cos(18x) dx
= ∫ sin^3(18x) (cos^2(18x))^2 cos(18x) dx
= ∫ sin^3(18x) (1 - sin^2(18x))^2 cos(18x) dx
= ∫ sin^3(18x) (1 + sin^4(18x) - 2sin^2(18x)) cos(18x) dx
let sin(18x) = u
18 cos(18x) dx = du
cos(18x) dx = du /18
1/18 ∫ u^3 (1 + u^4 - 2u^2) du
= 1/18 ∫ (u^3 + u^7 - 2u^5) du
= 1/18 [(1/4)u^4 + (1/8)u^8 - (1/3)u^6 ] + c
back substitute u = sin (18x)
(1/72)sin^4(18x) + (1/144)sin^8(18x) - (1/54)sin^6(18x) + c
Start with a substitution u = 18x
du = 18dx
∫sin³ (18x) cos^5 (18x) dx
= 1/18 * ∫sin³ (18x) cos^5 (18x) 18dx
= 1/18 * ∫sin³ (u) cos^5 (u) du
You seem to know what to do from there