[SOLVED] Indefinite Integral Problem

**fattydq** · April 5th 2009, 08:55 AM

I'm working on solving the integral of x/sqrt(2-x) and I'm using a trig substitution but the problem arises because normally 1-sin^2x=cos^2(x) but in this case the 2 is under a square root so I can't do that, leaving me stuck. COuld somebody help me out?

**skeeter** · April 5th 2009, 09:16 AM

Originally Posted by fattydq

I'm working on solving the integral of x/sqrt(2-x) and I'm using a trig substitution but the problem arises because normally 1-sin^2x=cos^2(x) but in this case the 2 is under a square root so I can't do that, leaving me stuck. COuld somebody help me out?

$u = 2-x<br />$

$du = -dx$

$x = 2 - u$

$-\int \frac{x}{\sqrt{2-x}} \, (-dx)$

$-\int \frac{2 - u}{\sqrt{u}} \, du$

$-\int \frac{2}{\sqrt{u}} - \sqrt{u} \, du$

integrate and back substitute.

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