
#1
Feb1512, 08:32 AM

P: 83

People, today i had a exam in math analysis and there was a integral to solve:
∫ x^2√(1x^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 ∫(1cos^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 = √(1x^2) dt/dx = x/(√(1x^2) ) So now i have a integral ∫(x^2*t*√(1x^2))/(x) dt ∫(x^2*t*t)/(x) dt ∫(x*t^2) dt as we know t = √(1x^2) so x= 1t^2 ∫((1t^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! 



#2
Feb1512, 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 I don't know how much that helps you though. 



#3
Feb1512, 09:08 AM

P: 83

omfg....
what a stupid error. Damn. ok i should do it by trigonometric substitution. 



#4
Feb1512, 01:20 PM

Sci Advisor
HW Helper
P: 4,301

Integral x^2(1x^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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