Integration using Trig. Substitution

[3.141592654]

Homework Statement

[tex]\int\sqrt{X^2+1}dX[/tex]

Homework Equations

The Attempt at a Solution

I used the substitution X=tan [tex]\theta[/tex]

So, [tex]dX=(sec^2 \theta[/tex]) d[tex]\theta[/tex]

Substituting in for X, I get:

[tex]\int\sqrt{(tan^2 \theta)+1}(sec^2 \theta) d\theta[/tex]

= [tex]\int\sqrt{(sec^2 \theta)}(sec^2 \theta) d\theta[/tex]

= [tex]\int(sec \theta)(sec^2 \theta) d\theta[/tex]

I then converted secants into cosines:

= [tex]\int\frac{1}{(cos \theta)(cos^2 \theta)} d\theta[/tex]

= [tex]\int\frac{1}{(cos \theta)(1-sin^2 \theta)} d\theta[/tex]

I then used U-sub:

[tex]u=sin \theta[/tex]

[tex]du=cos \theta[/tex]

[tex]\frac{du}{cos \theta}=d\theta[/tex]

= [tex]\int\frac{du}{(cos^2 \theta)(1-u^2)}[/tex]

= [tex]\int\frac{du}{(1-u^2)(1-u^2)}[/tex]

= [tex]\int\frac{du}{(1-2u^2+u^4)}[/tex]

= [tex]\int\frac{du}{(u^4-2u^2+1)}[/tex]

= [tex]\int\frac{du}{u^2(u^2-2)+1}[/tex]

I see this:

= [tex]\int\frac{du}{[u\sqrt{u^2-2}]^2+1}[/tex]

I was hoping I could then use substitution and have the integral of [tex]arctan[/tex], but it looks much more complicated than I thought. Am I anywhere near the right track?

 

[l'Hôpital]

[tex]
\int\frac{du}{(1-u^2)(1-u^2)}
[/tex]

At that point, you could attempt partial fractions. However, an easier way is here:

[tex]
\int(sec \theta)(sec^2 \theta) d\theta = \int sec^3 \theta d\theta
[/tex]

This is a famous (sorta, anyways) integral. Use parts twice.

Alternatively, since this integral is so famous, it even has its own wikipedia page.

http://en.wikipedia.org/wiki/Integral_of_secant_cubed

I'd suggest you trying the derivation yourself before looking at the "answer" but it's there if you need it.