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Question-76442




Question Number 76442 by aliesam last updated on 27/Dec/19
Answered by Tanmay chaudhury last updated on 28/Dec/19
x=tana    dx=sec^2 ada  sec^(−1) (((1+tan^2 a)/(1−tan^2 a)))→sec^(−1) (sec2a)=2a  ∫((2a)/(tana ×a))×sec^2 ada  2∫((d(tana))/(tana))→2ln(tana)  so  2∣lnx∣_e ^e^2  =so answer is 2(lne^2 −lne)=2(2−1)=2
x=tanadx=sec2adasec1(1+tan2a1tan2a)sec1(sec2a)=2a2atana×a×sec2ada2d(tana)tana2ln(tana)so2lnxee2=soansweris2(lne2lne)=2(21)=2
Commented by aliesam last updated on 28/Dec/19
god bless you sir
godblessyousir
Commented by Tanmay chaudhury last updated on 28/Dec/19
thank you sir
thankyousir

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