show that integral 1/x^2squareroot(x^2-a^2) =

((squareroot(x^2-a^2))/a^2x)+C using a trigonometric substitution

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- Oct 13th 2009, 04:07 AMmikegar813use trig substitution
show that integral 1/x^2squareroot(x^2-a^2) =

((squareroot(x^2-a^2))/a^2x)+C using a trigonometric substitution - Oct 13th 2009, 04:27 AMHallsofIvy
I remember that $\displaystyle sin^2(\theta)+ cos^2(\theta)= 1$ so that, dividing by $\displaystyle cos^2(\theta)$, $\displaystyle tan^2(\theta)- 1= sec^2(\theta)$ and then $\displaystyle a^2tan^2(\theta)- a^2= a^2sec^(\theta)$.

That suggests using the substitution $\displaystyle x= a tan(\theta)$ to get a "perfect square" inside the square root. - Oct 13th 2009, 06:10 AMtom@ballooncalculus
Typo... HoI surely means

dividing by $\displaystyle cos^2(\theta)$, $\displaystyle tan^2(\theta) + 1= sec^2(\theta)$ and then $\displaystyle a^2 \sec^2(\theta) - a^2 = a^2 \tan^2(\theta)$.

That suggests using the substitution $\displaystyle x= a \sec(\theta)$ to get a "perfect square" inside the square root

Edit:

Just in case a picture helps...

http://www.ballooncalculus.org/asy/t...ernal/aSec.png

... where

http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Carry on, anti-clockwise... g(theta) should be a nice simplification with a straightforward integral G. Then use a triangle or whatever to map back to x. Actually, as F(a sec theta) = G(theta) you can map from G...

http://www.ballooncalculus.org/asy/t...rnal/aSec1.png

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