How do you integrate #int dx/sqrt(9x^2+4)# using trig substitutions?

We want to find:

# I = int \ 1/sqrt(9x^2+4) \ dx #

For the integrand #1/sqrt(9x^2+4)# a suitable substitution should be:

# tanu = (3x)/2 => tan^2u = (9x^2)/4 #
# " "=> 9x^2=4tan^2u#

And differentiating wrt #x# we get:

# sec^2u(du)/dx=3/2 => 2/3sec^2u(du)/dx=1#

Applying the substitution along with #1+tan^2A-=sec^2A# we get;

# I = int \ (2/3sec^2u)/sqrt(4tan^2u+4) \ du #
# \ \ = 2/3 \ int \ (sec^2u)/sqrt(4(tan^2u+1)) \ du #
# \ \ = 2/3 \ int \ (sec^2u)/sqrt(4sec^2u) \ du #
# \ \ = 2/3 \ int \ (sec^2u)/(2secu) \ du #
# \ \ = 1/3 \ int \ secu \ du #
# \ \ = 1/3 \ int \ secu \ du #
# \ \ = 1/3 \ ln (secu + tanu)+c #

And again using #1+tan^2A-=sec^2A# we write

# sec^2u = 1+tan^2u #
# " " = 1+(9x^2)/4 #
# => sec u =sqrt(1+(9x^2)/4) #

And restoring the substituting we get:

# I = 1/3 \ ln (sqrt(1+(9x^2)/4) + (3x)/2)+c #

#int dx/sqrt(9x^2+4)#

Start with a simple arithmetic adjustment, as clearly we are looking to clean this up first.

#=1/2int dx/sqrt(9/4x^2+1)#

We then say that: #y^2 = 9/4 x^2 implies y = 3/2x# so # dy = 3/2 dx#.

#implies 1/3int dy/sqrt(y^2+1)#

Now we note that: #cosh^2 z - sinh^2 z = 1# so we can sub #y = sinh z# which means that: #dy = cosh z \ dz#

#implies 1/3int (cosh z \ dz)/sqrt(sinh^2 z+1)#

# = 1/3int dz#

#= 1/3 ( z + C)#

Now to reverse the sub's:

#= 1/3 sinh^(-1) y + C#, where C is arbitrary.

#= 1/3 sinh^(-1) (3/2 x) + C#

Now: # sinh^(-1) alpha= ln(alpha+sqrt (alpha^{2}+1})#

#implies 1/3 ( ln(3/2 x+sqrt ((3/2 x)^{2}+1}) )+ C#