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

1 Answer
Mar 10, 2018

#I = sqrt(x^2+ 1) - ln|(sqrt(x^2 + 1) + 1)/x| + C#

Explanation:

We let #x = tantheta#. Then #dx =sec^2theta d theta#.

#I = int sqrt(1 + tan^2theta)/tantheta * sec^2theta d theta#

#I = int sqrt(sec^2theta)/tantheta * sec^2theta d theta#

#I = int sec^3theta/tantheta d theta#

#I = int (1/cos^3theta)/(sintheta/costheta) d theta#

#I = int cscthetasec^2theta#

#I = int csctheta(1 + tan^2theta)d theta#

#I = int csctheta + cscthetatan^2theta d theta#

#I = int csctheta + secthetacostheta d theta#

Now these are two known integrals.

#I = sectheta - ln|csctheta + cottheta| + C#

IF #x/1 = tantheta#, then #sectheta = sqrt(x^2 + 1)# and #csctheta = sqrt(x^2 + 1)/x# and #cottheta = 1/x#.

#I = sqrt(x^2 + 1) - ln|sqrt(x^2 + 1)/x + 1/x| + C#

#I = sqrt(x^2+ 1) - ln|(sqrt(x^2 + 1) + 1)/x| + C#

Hopefully this helps!