Difficult integral for Trig Substitution

  • Thread starter bigred09
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ok i have been studying the in-depth processes of trigonometric substitution with integrals and this problem has me frusterated.

[tex]\int x^2\sqrt{(x^2-4)} dx[/tex]

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?
 
726
1
Try the substitution u = 2sec(theta)
 

HallsofIvy

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

[tex]\int x^2\sqrt{(x^2-4)} dx[/tex]
[itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] so [itex]sin^2(\theta)= 1- cos^2(\theta)[/itex] and, dividing on both sides by [itex]cos^2(\theta)[/itex], [itex]tan^2(\theta)= sec^2(\theta)- 1[/itex]. The substitution [itex]x= 2sec(\theta)[/itex], as bigred09 suggested, will reduce that squareroot to [itex]2 tan(\theta)[/itex].

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

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Alternatively, use the hyperbolic substitution [tex]x=2Cosh(u)[/tex]
 
ok well with trig substitution, i get

[tex]\int tan^2\theta sec^3\theta d\theta[/tex]

which doesn't help me. Can somone solve this integral then?
 
ok well with trig substitution, i get

[tex]\int tan^2\theta sec^3\theta d\theta[/tex]

which doesn't help me. Can somone solve this integral then?
Wrong! Look what Halls said, post #3.
 
Letting

[tex]x=2sec(\theta)=>4sec^2(\theta)\sqrt{4(sec^2(\theta)-1)}=4sec^2(\theta)*2\sqrt{tan^2(\theta)}=...{[/tex]
Edit: Disregard this!
 
Last edited:
726
1
@ sutupidmath:

You're forgetting about dx/d(theta)
 
right i actually forgot the coefficient 8 but that doesn't mess with the integral. and [tex]dx=sec\theta tan\theta[/tex]


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

[tex]8\int \frac{cos^5\theta}{sin^3\theta} d\theta[/tex]


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

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