How do I integrate (sin x)^6?

Letting z be the complex number cos x + i sin x, we will use the identities (zⁿ + 1/zⁿ) = 2 cos nx and (zⁿ - 1/zⁿ) = 2i sin nx. Both of these follow from De Moivre’s theorem and the odd and even properties of sine and cosine. To see this, notice that zⁿ + 1/zⁿ = cos nx + i sin nx + cos (-nx) + i sin (-nx) = cos nx + i sin nx + cos nx - i sin nx = 2 cos nx, and for the second identity, zⁿ - 1/zⁿ = cos nx + i sin nx - cos (-nx) - i sin (-nx) = cos nx + i sin nx - cos nx + i sin nx = 2i sin nx. We see that (2i sin x)⁶ = (z - 1/z)⁶ and    64 i⁶ sin⁶ x = z⁶ + 6(z)⁵(-1/z) + 15(z)⁴(-1/z)² + 20(z)³(-1/z)³ + 15(z)²(-1/z)⁴ + 6(z)(-1/z)⁵ + (-1/z)⁶= z⁶ - 6z⁴ + 15z² - 20 + 15/z² - 6/z⁴ +1/z⁶from the binomial theorem. Now, the method we use is to pair up the terms into groups that look like the identities above. Here, we are pairing the first with the last, the second with the second last, and so on, to give 64 i⁶ sin⁶ x = (z⁶ + 1/z⁶) - 6(z⁴+1/z⁴) + 15(z²+1/z²) - 20. Using the identity for cosine, we obtain 64 i⁶ sin⁶x = 2cos 6x - 12cos 4x + 30 cos 2x - 20, so 64 (-1)³ sin⁶x = 2cos 6x - 12cos 4x + 30 cos 2x - 20 and we can then see that sin⁶x = -1/32 cos 6x + 3/16 cos 4x - 15/32 cos 2x + 5/16. Finally, we can integrate both sides with respect to x, as now we can use recognised integrals on the right. ∫ sin⁶x dx = ∫ -1/32 cos 6x + 3/16 cos 4x - 15/32 cos 2x + 5/16 dx   = ∫ -1/32 cos 6x dx + ∫ 3/16 cos 4x dx - ∫ 15/32 cos 2x dx + ∫ 5/16 dx which we can evaluate by using ∫ cos nx dx = 1/n sin nx + C, to give our final answer, ∫ sin⁶x dx = -1/192 sin 6x + 3/64 sin 4x - 15/64 sin 2x + 5/16 x + C