`int (dx)/[(ax)^2 - b^2]^(3/2)` Evaluate the integral

`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`