How do you integrate #int 1/sqrt(x^2+1)# by trigonometric substitution?

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Answer 1 · 2016-08-14T19:36:04

#ln|x+sqrt(x^2+1)|+C#.

We take subst. #x=tant rArr dx= sec^2tdt#

Also, #1/sqrt(x^2+1)=1/sqrt(tan^t+1)=1/sect =cost#

Hence, #I=int1/sqrt(x^2+1)dt#

#=intcostsec^2tdt#

#=intsectdt#

#=ln|sect+tant|#

#=ln|x+sqrt(x^2+1)|+C#.