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First note that sin6(x)+cos6(x) is a sum of two cubes (sin2(x))3+(cos2(x))3 so it can be factored using a3+b3=(a+b)(a2−ab+b2) and the Pythagorean identity to get
sin6(x)+cos6(x)=(sin2(x))3+(cos2(x))3
=(sin2(x)+cos2(x))(sin4(x)−sin2(x)cos2(x)+cos4(x))
=sin4(x)−sin2(x)cos2(x)+cos4(x).
Therefore,
sin6x+cos6xsin2(x)cos2(x)=sin4(x)−sin2(x)cos2(x)+cos4(x)sin2(x)cos2(x)
=tan2(x)−1+cot2(x)
But tan2(x)=sec2(x)−1 and cot2(x)=csc2(x)−1. Hence,
sin6x+cos6xsin2(x)cos2(x)=sec2(x)+csc2(x)−3 and thus
∫sin6(x)+cos6(x)sin2(x)cos2(x) dx=∫sec2(x)+csc2(x)−3 dx
=tan(x)−cot(x)−3x+C
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