Integrals of powers of cos(x) - ∫cos^6(x)dx

Integrals of powers of trigonometric functions get increasingly difficult and tedious as the power increases. The idea is to reduce the power to the first order, and this can be done with power reduction formulas such as the half-angle formula. But as the power increases, the number of iterations of the half-angle formula also increases accordingly.

But there is a neat alternative for power reduction in complex numbers. Consider De Moivre's Theorem. It is probably the most efficient method of power reduction. For instance, in polar form...

z = cos(x) + isin(x)

If we raise z to the n^th power, then by De Moivre's Theorem...

zⁿ = [cos(x) + isin(x)]ⁿ = cos(nx) + isin(nx)

So we've reduced the power from n to 1 in just one step! Similarly...

1/z = z^-1 = [cos(x) + isin(x)]^-1 = cos(-x) + isin(-x) = cos(x) - isin(x)

1/zⁿ = z^-n = cos(-nx) + isin(-nx) = cos(nx) - isin(nx)

Now adding:

z + 1/z = 2cos(x)

zⁿ + 1/zⁿ = 2cos(nx)

These results are the key to performing integrals of larger powers of cosine.

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