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]

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]

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!

 

[JG89]

@ sutupidmath:

You're forgetting about dx/d(theta)

 

[sutupidmath]

@ sutupidmath:

You're forgetting about dx/d(theta)

 

[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?