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Question Number 118278 by Lordose last updated on 16/Oct/20

Evaluate  ∫_( 0) ^( (π/3)) tan^2 xsec((x/3))dx  ★

Evaluate0π3tan2xsec(x3)dx

Answered by MJS_new last updated on 16/Oct/20

∫tan^2  x sec (x/3) dx=       [t=csc (x/3) → dx=−3tan (x/3) sin (x/3)]  =−3∫(((3t^2 −4)^2 )/((t−2)^2 (t−1)^2 (t+1)^2 (t+2)^2 ))dt=  =−(4/3)∫(dt/((t−2)^2 ))+(2/9)∫(dt/(t−2))−(1/(12))∫(dt/((t−1)^2 ))+((35)/(36))∫(dt/(t−1))−     −(1/(12))∫(dt/((t+1)^2 ))−((35)/(36))∫(dt/(t+1))−(4/3)∫(dt/((t+2)^2 ))−(2/9)∫(dt/(t+2))=  ...  =((t(17t^2 −20))/(6(t^4 −5t^2 +4)))+(2/9)ln ∣((t−2)/(t+2))∣ +((35)/(36))ln ∣((t−1)/(t+1))∣  now put t=csc (x/3)  I don′t think we get a useable exact value.  I get ≈.713844

tan2xsecx3dx=[t=cscx3dx=3tanx3sinx3]=3(3t24)2(t2)2(t1)2(t+1)2(t+2)2dt==43dt(t2)2+29dtt2112dt(t1)2+3536dtt1112dt(t+1)23536dtt+143dt(t+2)229dtt+2=...=t(17t220)6(t45t2+4)+29lnt2t+2+3536lnt1t+1nowputt=cscx3Idontthinkwegetauseableexactvalue.Iget.713844

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