Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to integrate $$a\int_1^2 \frac{(t-1)^2-2t(t-1)}{t^2+(a(t-1)^2)^2} dt$$ $$=a\int_1^2 \frac{-t^2+1}{t^2+(a(t-1)^2)^2} dt $$ $$=-a\int_1^2 \frac{t^2-1}{t^2+(a(t-1)^2)^2} dt $$ with the hint that two trigonomic substitutions would be necessary and to consider using arctan. I have tried to tackle this a few different ways and get stuck every time. Can anyone help me start off?

share|improve this question
    
Maybe you can simplify this a bit first? –  ᴊ ᴀ s ᴏ ɴ Feb 27 '13 at 15:56
    
I think the key to the bottom is the two squares, so I'm wary of simplifying. –  Carly Feb 27 '13 at 15:58
    
Seems interessting, Mathematica needs more than minute (whithout a result) –  Dominic Michaelis Feb 27 '13 at 16:22
add comment

1 Answer

up vote 2 down vote accepted

Looks like we could make use of this $$ a\int \frac{-t^2+1}{t^2+(a(t-1)^2)^2} \; dt = -a \int \frac{1 - \frac{1}{t^2}}{1 + \left( a \left( \sqrt{t } - \frac{1}{\sqrt t } \right )^2 \right )^2 }dt = -a \int \frac{1 - \frac{1}{t^2}}{1 + \left( a \left( t + \frac{1}{ t } - 2 \right ) \right )^2 }dt $$

Substitute $\displaystyle a\left( t + \frac{1}{ t } - 2 \right ) = u$, you get $\displaystyle - \int \frac{ 1}{1 + (u)^2} du $

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.