Mooorre Integration by Substitution (1 Viewer)

Crosswinds · Jan 18, 2010

Okay so third and hopefully last dumb integration question:

∫ √(x/(1-x) dx {with limits between 1/2 and 0}, letting x = cos<sup>2</sup>@

Undoubtedly I'm doing something stupid again.

Please and thanks!

Cazic · Jan 18, 2010

$\text{We have } x = \cos^2(\theta) \Rightarrow dx = -2\sin(\theta)\cos(\theta) \, d\theta \ \text{ and } \ 1 - x = \sin^2(\theta).$

Now,

$\begin{align*}x = 0 &\Rightarrow \cos^2(\theta) = 0 \\ &\Rightarrow \theta = \frac{\pi}{2}, \ \text{ and} \\ x = \frac{1}{2} &\Rightarrow \cos(\theta) = \pm \frac{1}{\sqrt{2}} \\ &\Rightarrow \theta = \frac{\pi}{4},\end{align*}$

where the choice of the positive solution in the second set of implications is consistent with the positive square root in the original integrand.

Finally,

$\begin{align*}\int_{0}^{\frac{1}{2}} \sqrt{\frac{x}{1-x}} \, dx &= \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} -2 \cot(\theta) \sin(\theta)\cos(\theta) \, d\theta \\ &= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2(1-\sin^2(\theta)) \, d\theta & \text{(note the change of order of limits)}\\ &= 2\theta - \frac{2\theta}{2} + \frac{2\sin(2\theta)}{4} \ \biggr|_{\frac{\pi}{4}}^{\frac{\pi}{2}} \\ &= \frac{\pi}{4} - \frac{1}{2}.\end{align*}$

Drongoski · Jan 19, 2010

I think yr 2nd line ought to be: integral 2 cos^2 @ d@

Cazic · Jan 19, 2010

Drongoski said:
I think yr 2nd line ought to be: integral 2 cos^2 @ d@

Yep!

Crosswinds · Jan 19, 2010

Yaaaay awesome that makes sense! I think i just got confused how to integrate 2cos<sup>2</sup>@. Thank you!

Mooorre Integration by Substitution (1 Viewer)

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