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Question Number 124043 by liberty last updated on 30/Nov/20
∫sinxcosx1−2cos2xdx
Answered by john_santu last updated on 30/Nov/20
Answered by Dwaipayan Shikari last updated on 30/Nov/20
∫sinxcosx1−2cos2xdx1−2cos2x=t⇒4sin2x=dtdx=18∫dtt=18log(1−2cos2x)+C
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