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Ede4ka [16]
3 years ago
11

How do I evaluate the indefinite integral ∫sin(3x)⋅sin(6x)dxintsin(3x)*sin(6x)dx ?

Mathematics
1 answer:
ser-zykov [4K]3 years ago
5 0
If two exercices 
s sin(3x)dx - Ssin(6x)dx = -1/3cos(3 + 1/6cos(6x) + k
sin(3x) .sin(6x) = 1/2(cos(3x) - cos(9x) ) 
S= 1/2[ S cos(3x)dx - S cos(9x)dx ] = 1/2[1/3sin(3x) - 1/9sin(9x) ] + k


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