find the integral of :

(sin(x)-(1/3))(sin(x)-1)(sin(x)+1)

∫ (sin x – 1/3) (sin x – 1) (sin x + 1) dx = ∫ (sin x – 1/3) (sin^{2} x – 1) dx = – ∫ (sin x – 1/3) cos^{2} x dx =

–∫ sin x cos^{2} x dx + 1/3 ∫ cos^{2} x dx = the first integral is found by substitution

∫ cos^{2} x (d cos x) + 1/6 ∫ (1 + cos2 x) dx = (1/3) cos^{3} x + x/6 + (1/12) sin 2x + C

for this part ∫ (sin x – 1/3) (sin^{2} x – 1) dx how can you simplify it to (sin^{2} x – 1) I do not understand this

(sin x + 1)(sin x - 1) = sin^2 x - 1 by the formula of the difference of squares (a + b)(a - b) = a^2 - b^2, then use the Pythagorean identity sin^2 x + cos^2 x = 1.

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