First we remember that sinθ can be expressed in terms of powers of z, where z=cos(θ)+isin(θ), using the following:2isin(nθ)=zⁿ-z⁻ⁿ and 2cos(nθ)=zⁿ+z⁻ⁿ
so, [2isin(θ)]⁴=[z¹-z⁻¹]⁴ 16sin(θ)=(z)⁴(-z⁻¹)⁰+4(z)³(-z⁻¹)¹+6(z)²(-z⁻¹)²+4(z)¹(-z⁻¹)³+(z)⁰(-z⁻¹)⁴ by binomial exp.This simplifies to:16sin(θ)=(z⁴+z⁻⁴)-4(z²+z⁻²)+6but as we saw before (zⁿ+z⁻ⁿ)=2cos(nθ)so 16sin⁴(θ)=2cos(4θ)-8cos(2θ)+6so ∫sin⁴(x)=(1/16)∫2cos(4θ)-8cos(2θ)+6dx=3/8x-1/4sin(2x)+1/32sin(4x)+C.
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