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Question Number 34282 by math khazana by abdo last updated on 03/May/18
find∫π6π3dxcos(x)sin(x)
Commented by math khazana by abdo last updated on 07/May/18
I=∫π6π32dxsin(2x)=2∫π6π3dxsin(2x)=2[12ln∣tanx∣]π6π3=ln(tan(π3))−ln(tan(π6))=ln(3)−ln(13)=2ln(3)=ln(3).
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