Question

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

Answer 1

`intdx/(a^2x^2-b^2)^(3/2)=(-x)/(b^2sqrt(a^2x^2-b^2))+C`

Explanation:

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`