Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 18

intdx/[(ax)^2-b^2]^(3/2)
Let's use the integral substitution,
Let u=ax
du=adx
=>dx=(du)/a
=int(du)/(a(u^2-b^2)^(3/2))
=1/aint(du)/(u^2-b^2)^(3/2)
Now let's use the trigonometric substitution,
Let u=bsec(theta)
so du=bsec(theta)tan(theta)d theta
Plug these in the integrand,
=1/aint(bsec(theta)tan(theta))/(b^2sec^2(theta)-b^2)^(3/2)d theta
=1/aint(bsec(theta)tan(theta))/(b^2(sec^2(theta)-1))^(3/2)d theta
=1/aint(bsec(theta)tan(theta))/((b^2)^(3/2)(sec^2(theta)-1)^(3/2))d theta
Now use the identity:tan^2(theta)=sec^2(theta)-1
=1/aint(bsec(theta)tan(theta))/(b^3(tan^2(theta))^(3/2))d theta
=1/aint(sec(theta)tan(theta))/(b^2tan^3(theta))d theta
=1/(ab^2)intsec(theta)/(tan^2(theta))d theta
=1/(ab^2)int(1/cos(theta))/((sin^2(theta))/(cos^2(theta)))d theta
=1/(ab^2)int(1/cos(theta))*(cos^2(theta))/(sin^2(theta))d theta
=1/(ab^2)intcos(theta)/(sin^2(theta))d theta
Now let v=sin(theta)
=>dv=cos(theta)d theta
=1/(ab^2)int1/v^2dv
=1/(ab^2)(v^(-2+1)/(-2+1))
=1/(ab^2)(-1/v)
substitute back v=sin(theta)
=-1/(ab^2sin(theta))
We have used the substitution u=bsec(theta)
So,cos(theta)=b/u
using pythagorean identity,
sin^2(theta)+cos^2(theta)=1
sin^2(theta)+(b/u)^2=1
sin^2(theta)=1-b^2/u^2
sin^2(theta)=(u^2-b^2)/u^2
sin(theta)=sqrt(u^2-b^2)/u
Also recall we have used u=ax,
:.sin(theta)=sqrt((ax)^2-b^2)/(ax)
=-1/(ab^2sqrt((ax)^2-b^2)/(ax))
=(-1/(b^2))(x/sqrt((ax)^2-b^2))
Add a constant C to the solution,
=(-1/b^2)(x/sqrt((ax)^2-b^2))+C