How do you integrate #int1/((ax)^2-b^2)^(3/2)# by trigonometric substitution?

This can be written as:

#intdx/(a^2x^2-b^2)^(3/2)#

We will use the substitution #x=b/asectheta#. This also implies that #dx=b/asecthetatanthetad theta#. Substituting, this gives us:

#=int(b/asecthetatanthetad theta)/(a^2(b^2/a^2sec^2theta)-b^2)^(3/2)#

#=int(bsecthetatanthetad theta)/(a(b^2sec^2theta-b^2)^(3/2))#

#=int(bsecthetatanthetad theta)/(a(b^2)^(3/2)(sec^2theta-1)^(3/2))#

Since #1+tan^2theta=sec^2theta#, so #sec^2theta-1=tan^2theta#:

#=int(bsecthetatanthetad theta)/(ab^3(tan^2theta)^(3/2))#

#=int(secthetatanthetad theta)/(ab^2tan^3theta)#

#=1/(ab^2)int(secthetad theta)/tan^2theta#

#=1/(ab^2)int1/costheta(cos^2theta/sin^2theta)d theta#

#=1/(ab^2)intcostheta/sin^2thetad theta#

Let #u=sintheta# so #du=costhetad theta#. This gives us:

#=1/(ab^2)int(du)/u^2#

Integrating #u^-2# using the power rule for integration, which gives #u^-1/(-1)=-1/u#:

#=1/(ab^2)(-1/u)#

Since #u=sintheta#:

#=-1/(ab^2)csctheta#

Since #x=b/asectheta#, we see that #theta="arcsec"((ax)/b)#:

#=-1/(ab^2)csc("arcsec"((ax)/b))#

To find #csc("arcsec"((ax)/b))#, we want to find the cosecant of a triangle where the secant is #(ax)/b#.

Since secant is the reciprocal of cosine, we see that #ax# is the hypotenuse is #b# is the adjacent side. Through the Pythagorean theorem, we see that the opposite side is #sqrt(a^2x^2-b^2)#.

This gives us a cosecant, which as the reciprocal of sine is the hypotenuse over the opposite side, or #(ax)/sqrt(a^2x^2-b^2)#.

#=-1/(ab^2)((ax)/sqrt(a^2x^2-b^2))#

#=(-x)/(b^2sqrt(a^2x^2-b^2))+C#