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Question Number 64332 by Chi Mes Try last updated on 16/Jul/19
∫π/3π/61sin2xdx=
Commented by mathmax by abdo last updated on 16/Jul/19
letA=∫π6π3dxsin(2x)changement2x=tgiveA=∫π32π3dt2sint=tan(t2)=u12∫1332du(1+u2)2u1+u2=12∫133duu=12[ln∣u∣]133=12{ln(3)+ln(3)}=ln(3)2A=ln(3)2.
Commented by Tony Lin last updated on 17/Jul/19
∫π6π31sin2xdx=∫π6π3csc2xdx=−12ln∣csc2x+cot2x∣∣π6π3=−12ln∣cotx∣π6π3=−ln(13)2+ln(3)2=ln32
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