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

2 Answers
Dec 22, 2016

#sqrt(x^2 + 1) + C#

Explanation:

Let #x= tantheta#. Then #dx = sec^2thetad theta#

#=> int tan theta/sqrt((tan^2theta + 1)) sec^2theta d theta#

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

#=> int tantheta/sectheta sec^2theta d theta#

#=> int tan theta sec theta d theta#

This is a common integral--#int(tanxsecx)dx = secx + C#.

#=> sec theta + C#

We now draw an imaginary triangle.

enter image source here

The definition of #sectheta# is #"hypotenuse"/("side adjacent" theta)# because #secx = 1/cosx#. In this image, #sectheta = sqrt(x^2 + 1)#.

Therefore, the integral can be simplified to #sqrt(x^2 + 1) + C#.

Hopefully this helps!

Dec 23, 2016

By inspection rather than trig substitution.

Explanation:

Notice that the #x# in the top is proportional to the derivative of the function under the square root. So just write down #(x^2+1)^(-1/2+1)# as a trial (rather like integrating any power of #x#) and differentiate it. By the chain rule, the function under the square root will provide the #2x# and the #2#'s will cancel, so the trial was an immediate success.

Alternatively substitute #u=x^2+1#, #dx=(du)/(2sqrt(u-1))# and the integral becomes #(1/2)int u^(-1/2)du# = #u^(1/2)+C#=#sqrt(x^2+1)+C#