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Integrate cos(3+2x)sin(1+2x) dx my sol: [(cos3)cos2x-(sin3)sin2x][sin1cos2x + sin2x cos1] dx 0.5cos3sin1(1+cos4x) + 0.5cos3cos1sin2x dsin2x - 0.5sin3sin1sin2x dsin2x-0.5sin3cos1(1-cos4x) dx 0.5cos3sin1(x+0.25sin4x)+ 0.5(0.5 cos3cos1 - 0.5 sin3sin1)sin^2(2x)-0.5 sin3cos1(x-0.25sin4x) + c Any easier solution?
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Hint : \[2\cos A\sin B = \sin(A+B) - \sin(A-B)\]
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