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Question Number 118278 by Lordose last updated on 16/Oct/20
Evaluate∫0π3tan2xsec(x3)dx★
Answered by MJS_new last updated on 16/Oct/20
∫tan2xsecx3dx=[t=cscx3→dx=−3tanx3sinx3]=−3∫(3t2−4)2(t−2)2(t−1)2(t+1)2(t+2)2dt==−43∫dt(t−2)2+29∫dtt−2−112∫dt(t−1)2+3536∫dtt−1−−112∫dt(t+1)2−3536∫dtt+1−43∫dt(t+2)2−29∫dtt+2=...=t(17t2−20)6(t4−5t2+4)+29ln∣t−2t+2∣+3536ln∣t−1t+1∣nowputt=cscx3Idon′tthinkwegetauseableexactvalue.Iget≈.713844
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