How do you evaluate #\int \frac { - d t } { \sqrt { t ^ { 2} + 1^ { 2} } }#?

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Answer 1 · 2017-12-07T16:42:07

#-ln(|t+sqrt(t^2+1)|)+C#

#int(-dt)/(sqrt(t^2+1)#

#=-int(dt)/(sqrt(t^2+1))---(1)#

#t=tanu=>dt=sec^2udu#

#(1)rarr-int(sec^2udu)/sqrt(tan^2u+1---(2)#

but# sqrt(tan^2u+1)=sqrt(sec^2u)=secu#

#(2)rarr-intsec^2u/secudu#

#=-intsecudu#

#=-intsecuduxx(secu+tanu)/(secu+tanu)#

#=-int(sec^2u+secutanu)/((secu+tanu))du----(3)#

#d/(du)(secu+tanu)=secutanu+sec^2u#

the integral is of the form

#int(f'(x))/(f(x))dx=ln|(f(x)|+C#

#(3)rarr-ln|(secu+tanu)|+C---(4)#

#because t=tanu, secu=sqrt(1+x^2)#

#(4)rarr-ln(|t+sqrt(t^2+1)|)+C#

Answer 2 · 2017-12-07T18:11:47

#-Ln(sqrt(t^2+1)+t)+C=Ln(sqrt(t^2+1)-t)+C#

#int -(dt)/sqrt(t^2+1)#

After using #t=sinhu# and #dt=coshu*du# transforms, this integral became,

#int -(coshu*du)/coshu#

=#int -du#

=#-u+C#

After using #t=sinhu=(e^u-e^(-u))/2# and #u=ln(sqrt(t^2+1)+t)# inverse transforms, I found

#int -(dt)/sqrt(t^2+1)#

=#-Ln(sqrt(t^2+1)+t)+C#

=#Ln[1/(sqrt(t^2+1)+t)]+C#

=#Ln(sqrt(t^2+1)-t)+C#