How do you integrate #int 1/sqrt(4x+8sqrtx+12) # using trigonometric substitution?

1 Answer
Sep 7, 2016

#sqrt(x+2sqrtx+3)-ln(abs(sqrtx+sqrt(x+2sqrtx+3)+1))+C#

Explanation:

We have:

#intdx/sqrt(4x+8sqrtx+12)#

Factor #sqrt4=2# from the denominator:

#=1/2intdx/sqrt(x+2sqrtx+3)#

Complete the square in the denominator. Think about it in terms of #x^2+2x+3#, and then switch #x# terms to #sqrtx# and #x^2# to #x#:

#=1/2intdx/sqrt((sqrtx+1)^2+2)#

Now, let #sqrtx+1=sqrt2tantheta#. Note that this implies that #dx/(2sqrtx)=sqrt2sec^2thetad theta#.

Rearranging some:

#=int(sqrtxdx)/(2sqrtxsqrt((sqrtx+1)^2+2))#

Now, performing the substitutions. We have #dx/(2sqrtx)=sqrt2sec^2thetad theta# present, as well as #sqrtx=sqrt2tantheta-1# in the numerator. Don't forget the switch in the radical in the denominator either:

#=int((sqrt2tantheta-1)sqrt2sec^2thetad theta)/sqrt(2tan^2theta+2)#

Factoring #sqrt2# from the denominator and using #sqrt(tan^2theta+1)=sectheta#:

#=int((sqrt2tantheta-1)sqrt2sec^2theta)/(sqrt2sectheta)d theta#

#=int(sqrt2tantheta-1)secthetad theta#

#=sqrt2inttanthetasecthetad theta-intsecthetad theta#

Both of which are common integrals:

#=sqrt2sectheta-ln(abs(sectheta+tantheta))#

Write #sectheta# in terms of #tantheta#:

#=sqrt2sqrt(tan^2theta+1)-ln(abs(sqrt(tan^2theta+1)+tantheta))#

Using #tantheta=(sqrtx+1)/sqrt2#:

#=sqrt2sqrt((sqrtx+1)^2/2+1)-ln(abs(sqrt((sqrtx+1)^2/2+1)+(sqrtx+1)/sqrt2))#

#=sqrt2sqrt(((sqrtx+1)^2+2)/2)-ln(abs(sqrt(((sqrtx+1)^2+2)/2)+(sqrtx+1)/sqrt2))#

#=sqrt((sqrtx+1)^2+2)-ln(abs((sqrt((sqrtx+1)^2+2)+sqrtx+1)/sqrt2))#

Note that the #1/sqrt2# can actually be taken from the denominator of the logarithm as #-lnsqrt2#, which will be absorbed into the constant of integration in the next step:

#=sqrt(x+2sqrtx+3)-ln(abs(sqrtx+sqrt(x+2sqrtx+3)+1))+C#