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Question Number 187528 by sciencestudentW last updated on 18/Feb/23
∫7π1.59π2dx1−sinx=?
Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23
I=∫dx1−sinx=∫dx∣cosx2−sinx2∣=12∫dxcos(x2+π4)=12ln∣sec(x2+π4)+tan(x2+π4)∣+C
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