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Question Number 141532 by sarkor last updated on 20/May/21
Answered by qaz last updated on 20/May/21
∫3sinx−2cosx1+cosxdx=∫6sinx2cosx2−4cos2x2+22cos2x2dx=∫(3tanx2−2+sec2x2)dx=6ln∣cosx2∣−2x+2tanx2+C
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