# Difficult integral for Trig Substitution

#### 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?

#### JG89

Try the substitution u = 2sec(theta)

#### HallsofIvy

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

$$\int x^2\sqrt{(x^2-4)} dx$$
$sin^2(\theta)+ cos^2(\theta)= 1$ 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 $2 tan(\theta)$.

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?

#### arildno

Homework Helper
Gold Member
Dearly Missed
Alternatively, use the hyperbolic substitution $$x=2Cosh(u)$$

#### 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?

#### sutupidmath

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?
Wrong! Look what Halls said, post #3.

#### 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:

@ sutupidmath:

#### 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?

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