How do you integrate #int 1/sqrt(4x^2+16x+8) # using trigonometric substitution?

2 Answers
Jun 16, 2018

#1/2ln|x+2+sqrt(x^2+4x+2)|+C#

Explanation:

It is in general a good idea to divide any polynomial that we have to work with through by its leading coefficient; this almost always makes it simpler to work with. So:

#int1/sqrt(4x^2+16x+8)dx=1/2int1/sqrt(x^2+4x+2)dx#

Complete the square:

#1/2int1/sqrt((x+2)^2-2)dx#

Make a substitution to cause the square root to vanish via a trig identity. In this case we'll use #x+2=sqrt(2)sectheta#. We could also use #csc# (which would be similar) or, straying into hyperbolic functions, #cosh# (which would be easier).

As #x=-2+sqrt(2)sectheta#, #dx/(d theta)=sqrt(2)secthetatantheta#, and so our integral becomes

#1/2int1/sqrt(2sec^2theta-2)*sqrt(2)secthetatanthetad theta#

Recalling the identity #tan^2theta=sec^2theta-1#, we obtain

#1/2int1/(sqrt(2)tantheta)*sqrt(2)secthetatanthetad theta#

which simplifies considerably to become

#1/2intsectheta d theta#

The integral of #secx# is #ln|secx+tanx|#. Two demonstrations of this are given here: https://www.math.ubc.ca/~feldman/m121/secx.pdf
So we get

#1/2ln|sectheta+tantheta|+C#

Transform back to our original variable, #x#:

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

Rearrange a little to obtain a simpler form:

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

Note that the #-1/2lnsqrt(2)# term is a constant and can therefore be folded into the integration constant #C#

#1/2ln|x+2+sqrt(x^2+4x+2)|+C#

If we wished to write this in the same style as the original question formula, we could use the same observation about the integration constant to add #1/2ln2# without changing the answer:

#1/2ln|x+2+sqrt(x^2+4x+2)|+C#
#1/2ln|x+2+sqrt(x^2+4x+2)|+1/2ln2+C#
#1/2ln|2(x+2)+sqrt(4x^2+16x+8)|+C#

It's arguable whether it's better to write it so that the expression inside the root matches that in the original integral or so that it is a simpler expression. I think I prefer our first answer myself:

#1/2ln|x+2+sqrt(x^2+4x+2)|+C#

It's more elegant.

Jun 16, 2018

#I=1/2ln|(x+2)+sqrt(x^2+4x+2)|+C#

Explanation:

Here,

#I=int1/sqrt(4x^2+16x+8)dx#

#I=int1/(2sqrt(x^2 +4x+2))dx#

#I=1/2int1/sqrt(x^2+4x+4-2)dx#

#I=1/2int1/sqrt((x+2)^2-(sqrt2)^2)dx#

Subst. #x+2=sqrt2secu=>dx=sqrt2secutanudu#

#=>secu=(x+2)/sqrt2#

#I=1/2int1/sqrt(2sec^2u-2)xxsqrt2secutanudu#

#:.I=1/2int(sqrt2secutanu)/(sqrt2tanu)du#

#=>=1/2intsecudu#

#:.I=1/2ln|secu+tanu|+c#

#=1/2ln|secu+sqrt(sec^2u-1)|+c#

Subst. back , #secu=(x+2)/sqrt2#

#I=1/2ln|(x+2)/sqrt2+sqrt((x+2)^2/2-1)|+c#

#I=1/2ln|(x+2)/sqrt2+sqrt((x+2)^2-2)/sqrt2|+c#

#I=1/2ln|((x+2)+sqrt(x^2+4x+2))/sqrt2|+c#

#I=1/2ln|(x+2)+sqrt(x^2+4x+2)|-1/2lnsqrt2+c#

#I=1/2ln|(x+2)+sqrt(x^2+4x+2)|+C,#
# where,C=c-1/2lnsqrt2#