Not sure how to solve this integral

[ku1005]

Hi, with the follwing integral, im not really sure how to go about solving it. I was thinking along the lines of substitution, but I cant see anyhting that will work. I also was thinking of Trig subs, but it doesnt really mimic any of the well known trig intergals (well at least the ones I know)....any help/tips is appreciated!!

I have to solve this exactly



cheers

 

[tkhunny]

\(\displaystyle \L\;u^{2} = x^{2} - 9\) seems useful.

Simplifying the expression with the worse exponent often is a good idea.

 

[ku1005]

how does that help but???....because it still doesnt get rid of the x^5??? sorry i dont follow, I already did try using u = sqrt(x^2 - 9 ).....or as you say

u^2 = x^2 - 9

but I cant see how the derivative of this can be substituted and used to cancel the x^5 ???

thanks

 

[Opalg]

I already did try using u = sqrt(x^2 - 9 ).....or as you say

u^2 = x^2 - 9

but I cant see how the derivative of this can be substituted and used to cancel the x^5 ???

If u^2 = x^2 - 9 then u du = x dx, and \(\displaystyle \displaystyle\int x^5(x^2-9)^{3/2}\,dx = \int x^4(x^2-9)^{3/2}\,x\,dx = \int(u^2+9)^2u^3\,u\,du.\) Multiply out the square and you just have a polynomial to integrate.

 

[tkhunny]

because it still doesnt get rid of the x^5???

That makes me nervous. "It" doesn't do anything. You have to do it.

u = sqrt(x^2 - 9 ).....or as you say u^2 = x^2 - 9

Defintely NOT the same thing. The idea is to make it simpler, not more complicated.

It's okay to have chunks that are not covered with the calculation of du. Use your substitution to continue transforming the integrand, as was demonstrated nicely by Opalg. You need one x for the du-part. The x^4 is a simple solution from the original substitution.

 

[ku1005]

thanks very much Opalg and Tkhunny.....i now understand...much appreciated!

 

[mark]

you can use trig substitution

you have x^2 – a^2

with a equal to 3

so, x = 3sin(theta)