# Difficult integral for Trig Substitution

1. Mar 20, 2009

### bigred09

ok i have been studying the in-depth processes of trigonometric substitution with integrals and this problem has me frusterated.

$$\int x^2\sqrt{(x^2-4)} dx$$

The evaluation is clear (from an old Table of Integrals I found), but the derivation is not at all clear, which is what i want to know.

I also tried to solve this by integration by parts, but every approach ended with an even more complicated integral, so trig substitution is probably the best choice.

Can anyone help?

2. Mar 20, 2009

### JG89

Try the substitution u = 2sec(theta)

3. Mar 21, 2009

### HallsofIvy

$sin^2(\theta)+ cos^2(\theta)= 1$ so $sin^2(\theta)= 1- cos^2(\theta)$ and, dividing on both sides by $cos^2(\theta)$, $tan^2(\theta)= sec^2(\theta)- 1$. The substitution $x= 2sec(\theta)$, as bigred09 suggested, will reduce that squareroot to $2 tan(\theta)$.

4. Mar 21, 2009

### arildno

Alternatively, use the hyperbolic substitution $$x=2Cosh(u)$$

5. Mar 22, 2009

### bigred09

ok well with trig substitution, i get

$$\int tan^2\theta sec^3\theta d\theta$$

which doesn't help me. Can somone solve this integral then?

6. Mar 22, 2009

### sutupidmath

Wrong! Look what Halls said, post #3.

7. Mar 22, 2009

### sutupidmath

Letting

$$x=2sec(\theta)=>4sec^2(\theta)\sqrt{4(sec^2(\theta)-1)}=4sec^2(\theta)*2\sqrt{tan^2(\theta)}=...{$$
Edit: Disregard this!

Last edited: Mar 23, 2009
8. Mar 23, 2009

### JG89

@ sutupidmath:

9. Mar 23, 2009

### sutupidmath

10. Mar 27, 2009

### bigred09

right i actually forgot the coefficient 8 but that doesn't mess with the integral. and $$dx=sec\theta tan\theta$$

so what halls said wass valid. all i did was simplify it more. even more so it looks like this:

$$8\int \frac{cos^5\theta}{sin^3\theta} d\theta$$

so uh...seriously...any ideas on solving this?