How do you use substitution to integrate #2/(xsqrt(4lnx- (lnx)^2 ))#?

1 Answer
Jul 23, 2015

The answer turned out to be #4arcsin(sqrt(lnx)/2) + C#. Just a note; Wolfram Alpha gives a much more complicated answer, and it's not particularly nice of a result when, for example, you are typing it in for an online homework problem.

Wolfram Alpha gives its alternative answer as:

#= (4sqrt(lnx - 4)sqrt(lnx)ln(sqrt(lnx - 4) + sqrt(lnx)))/sqrt(-(lnx - 4)lnx)#

#= (4sqrt(lnx - 4)sqrt(lnx)ln(sqrt(lnx - 4) + sqrt(lnx)))/sqrt(4lnx - ln^2x)#

which... let's face it, looks terrible! :)


I feel like trig substitution should be done somewhere here... Let's say for now...

Let:
#u = lnx#
#du = 1/xdx#

#=> int 2/(xsqrt(4u - u^2))dx#

#= 2int 1/(sqrt(4u - u^2))du#

Okay, so this looks a little better. What if we factored something out?

#= 2int 1/(sqrtusqrt(4 - u))du#

#= 2int 1/(sqrtusqrt(2^2 - (sqrtu)^2))du#

Now, what if we said...
#sqrtu = 2sintheta#
#u = 4sin^2theta#
#du = 8sinthetacosthetad theta#
#sqrt(4 - u) = sqrt(2^2 - 2^2sin^2theta) = 2costheta#

More manageable now.

#= 2int 1/(cancel(2)cancel(sintheta)cancel(2)cancel(costheta))*cancel(8)^2cancel(sinthetacostheta)d theta#

#= 4int d theta#

Huh? How... easy?

#= 4theta#

Now, we had #2sintheta = sqrtu#, so:

#theta = arcsin((sqrtu)/2)#

Thus:

#theta = arcsin(sqrt(lnx)/2)#

And so we have:

#= color(blue)(4arcsin(sqrt(lnx)/2) + C)#

That is our answer. It seems rather intuitive considering the resemblance of the original with #1/(sqrt(1-u^2))#.

Let's try to differentiate this and see if we get back to the original.

#(dy)/(dx)[arcsinu] = 1/sqrt(1-u^2)((du)/(dx))#

Thus, with #u = sqrt(lnx)/2#, we have, after some Chain Rule action...

Cancel out terms:
#4(d)/(dx)[arcsin(sqrt(lnx)/2)] = cancel(4)*1/(sqrt(1-((sqrtlnx)/2)^2))*1/cancel(2)*1/(cancel(2)sqrt(lnx))*1/x#

Shift values around:
#= 1/(xsqrt(lnx)sqrt(1-((sqrtlnx)/2)^2))#

Multiply out the square:
#= 1/(xsqrt(lnx)sqrt(1-(lnx)/4))#

Distribute into the square root:
#= 1/(xsqrt(lnx-(lnx)^2/4))#

Factor #sqrt(1/4)# out:
#= 1/(xsqrt(1/4)sqrt(4lnx-(lnx)^2))#

Shift that around:
#= sqrt4/(xsqrt(4lnx-(lnx)^2))#

Done!
#= color(green)(2/(xsqrt(4lnx-(lnx)^2)))#