By Angle-Addition, cos(2x) = 2cos(x)^2-1
So, sin(x)cos(2x) = [2cos(x)^2-1]sin(x) = 2sin(x)cos(x)^2 - sin(x)
Int[2sin(x)cos(x)^2 - sin(x)] = (-2/3)cos(x)^3 + cos(x) +K
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use the double angle formula for cos(2x) which is: cos(2x)=2cos^2(x)-1 by this relation cos^2(x)=(cos(2x)+1)/2 now we'd integrate this instead this will give sin(2x)/4+x/2 =) hope this helps
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Cos(2x) = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x).Source: ChaCha.com
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
2 x cosine squared x -1 which also equals cos (2x)