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

Question

Emilia Aguilar · Accepted Answer

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

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#

Emilia Aguilar · Answer

undefined The antiderivative of $\frac{\sec^2(x) \tan^2(x)}{\sec^2(x) - R^2}$ with respect to $x$ is:

$$\frac{1}{2} \left(\ln\left|\frac{\sec(x) + \tan(x)}{\sec(x) - \tan(x)}\right| - R^2 \ln\left|\sec(x) - R\right| + R^2 \ln\left|\sec(x) + R\right|\right) + C$$

where $C$ is the constant of integration.