Question

What is the antiderivative of `(sec(x)^2)(tan(x)^2)/((sec(x)^2-(R^2))` where R is a constant?

Answer 1

`tanx-lnabs(sec^2x-R^2)/(2sqrt(R^2-1))+C`

Explanation:

For simplicity's sake, we will use the notation `sec(x)^2=sec^2x`.

We have:

`I=int(sec^2xtan^2x)/(sec^2x-R^2)dx`

In the denominator, let `sec^2x=tan^2x+1`:

`I=int(sec^2xtan^2x)/(tan^2x+1-R^2)dx`

Let `u=tanx` so `du=sec^2xdx`:

`I=intu^2/(u^2+1-R^2)du`

Rewriting the numerator:

`I=int((u^2+1-R^2)-1+R^2)/(u^2+1-R^2)du`

`I=int(1+(R^2-1)/(u^2+1-R^2))du`

Integrating `1` and bring `R^2-1` from the integral:

`I=u+(R^2-1)int1/(u^2+1-R^2)du`

We will now perform partial fraction decomposition on the remaining integrand:

`1/(u^2+1-R^2)=1/(u^2-(R^2-1))=1/((u+sqrt(R^2-1))(u-sqrt(R^2-1)))`

So we will use the decomposition:

`1/((u+sqrt(R^2-1))(u-sqrt(R^2-1)))=A/(u+sqrt(R^2-1))+B/(u-sqrt(R^2-1))`

`1=A(u-sqrt(R^2-1))+B(u+sqrt(R^2-1))`

Letting `u=sqrt(R^2-1)`:

`1=A(sqrt(R^2-1)-sqrt(R^2-1))+B(sqrt(R^2-1)+sqrt(R^2-1))`

`1=B(2sqrt(R^2-1))`

`B=1/(2sqrt(R^2-1))`

Letting `u=-sqrt(R^2-1)`:

`1=A(-sqrt(R^2-1)-sqrt(R^2-1))+B(-sqrt(R^2-1)+sqrt(R^2-1))`

`1=A(-2sqrt(R^2-1))`

`A=(-1)/(2sqrt(R^2-1))`

Then:

`1/(u^2+1-R^2)=(-1)/(2sqrt(R^2-1)(u+sqrt(R^2-1)))+1/(2sqrt(R^2-1)(u-sqrt(R^2-1)))`

Returning to the integral:

`I=u-1/(2sqrt(R^2-1))int1/(u+sqrt(R^2-1))du+1/(2sqrt(R^2-1))int1/(u-sqrt(R^2-1))du`

These are both deceptively simple integrals, since `sqrt(R^2-1)` is a constant. The first integral can be integrated with the substitution `v=u+sqrt(R^2-1)` so `du=dv`, giving `int1/vdv=lnabsv`.

`I=u-1/(2sqrt(R^2-1))lnabs(u+sqrt(R^2-1))+1/(2sqrt(R^2-1))lnabs(u-sqrt(R^2+1))`

From `u=tanx`:

`I=tanx-(lnabs(tanx+sqrt(R^2-1))+lnabs(tanx-sqrt(R^2-1)))/(2sqrt(R^2-1))`

Using `ln(a)+ln(b)=ln(ab)`:

`I=tanx-lnabs(tan^2x-(R^2-1))/(2sqrt(R^2-1))`

Using `tan^2x+1=sec^2x`:

`I=tanx-lnabs(sec^2x-R^2)/(2sqrt(R^2-1))+C`