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28 September, 17:09

How do I evaluate the indefinite integral

∫sin (3x) ⋅sin (6x) dxintsin (3x) * sin (6x) dx?

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  1. 28 September, 17:24
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    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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