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Question Number 167490 by mathlove last updated on 18/Mar/22
∫sin3xcos2xdx=?
Answered by nimnim last updated on 18/Mar/22
I=∫sin2cos2xsinxdx=∫(1−cos2x)cos2xsinxdx=∫cos2xsinxdx−∫cos4xsinxdx=−13cos3x+15cos5x+C
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