Integral x^2(1-x^2)


by Fabio010
Tags: integral, x21x2
Fabio010
Fabio010 is offline
#1
Feb15-12, 08:32 AM
P: 83
People, today i had a exam in math analysis and there was a integral to solve:

∫ x^2√(1-x^2) dx

ok, i started to think about the trigonometric substitution. x= sint

but, with that substitution now i have a ∫sin^2tcos^2t dt

so i have to do something like ∫(1-cos^2t)(cos^2t) dt ok and i thought (no thanks...)
i never learned how to solve a integral with the trigonometric formula, so solve something like
∫cos^4t dt takes a lot of time.


So i tried t = √(1-x^2)

dt/dx = -x/(√(1-x^2) )

So now i have a integral

-∫(x^2*t*√(1-x^2))/(x) dt
-∫(x^2*t*t)/(x) dt
-∫(x*t^2) dt

as we know t = √(1-x^2) so x= 1-t^2

-∫((1-t^2)t^2) dt = -∫t^2 - t^4 dt

ok now it is easy...

Please tell me that i did it in the correct way!
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CompuChip
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#2
Feb15-12, 09:00 AM
Sci Advisor
HW Helper
P: 4,301
You can easily check by differentiating, however I think that you'll find you're off by a factor of x then.

The problem seems to be
as we know t = √(1-x^2) so x= 1-t^2
if t2 = 1 - x2 then x2 = 1 - t2.
I don't know how much that helps you though.
Fabio010
Fabio010 is offline
#3
Feb15-12, 09:08 AM
P: 83
omfg....

what a stupid error.

Damn. ok i should do it by trigonometric substitution.

CompuChip
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#4
Feb15-12, 01:20 PM
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P: 4,301

Integral x^2(1-x^2)


I just made the observation that
[tex]x^2 \sqrt{1 - x^2} \propto x \cdot \frac{d}{dx} (1 - x^2)^{3/2}[/tex]
so maybe you can try partial integration.


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