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Question Number 64332 by Chi Mes Try last updated on 16/Jul/19

∫_(π/6) ^(π/3)   (1/(sin 2x)) dx =

π/3π/61sin2xdx=

Commented by mathmax by abdo last updated on 16/Jul/19

let A =∫_(π/6) ^(π/3)  (dx/(sin(2x))) changement 2x=t give   A =∫_(π/3) ^((2π)/3)    (dt/(2sint)) =_(tan((t/2))=u)   (1/2) ∫_(1/(√3)) ^(√3)    ((2du)/((1+u^2 )((2u)/(1+u^2 )))) =(1/2) ∫_(1/(√3)) ^(√3)  (du/u)  =(1/2)[ln∣u∣]_(1/(√3)) ^(√3)   =(1/2){ln((√3))+ln((√3))} =((ln(3))/2)  A =((ln(3))/2) .

letA=π6π3dxsin(2x)changement2x=tgiveA=π32π3dt2sint=tan(t2)=u121332du(1+u2)2u1+u2=12133duu=12[lnu]133=12{ln(3)+ln(3)}=ln(3)2A=ln(3)2.

Commented by Tony Lin last updated on 17/Jul/19

∫_(π/6) ^(π/3) (1/(sin2x))dx  =∫_(π/6) ^(π/3) csc2xdx  =−(1/2)ln∣csc2x+cot2x∣∣_(π/6) ^(π/3)   =−(1/2)ln∣cotx∣_(π/6) ^(π/3)   =−((ln((1/(√3))))/2)+((ln((√3)))/2)  =((ln3)/2)

π6π31sin2xdx=π6π3csc2xdx=12lncsc2x+cot2xπ6π3=12lncotxπ6π3=ln(13)2+ln(3)2=ln32

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