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I'm taking this class where I often have to calculate integrals of the form $$ \int\sqrt{\alpha-x^2}dx, ~~~\alpha \in \mathbb{R}$$

Yes you can calculate these by trigonometric substitution or hyperbolic substitution, but the process is lengthy and the result is always a relatively big expression, so I find trig substitution to be annoying. So I ask: is there some relatively easier method for calculating integrals like this? I have heard of the "differentiation under the sign" method of integration and people always talk about it like it works miracles. Can it be applied here and if it can, would it be easier than trig substitution?

Thanks.

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    $\begingroup$ Trigonometric or hyperbolic substitutions are easiest ways to do this. $\endgroup$ – Nosrati May 26 at 1:05
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    $\begingroup$ Is is $\alpha$ or $\alpha^2$ inside the sqrt ? $\endgroup$ – Ak19 May 26 at 1:11
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    $\begingroup$ You can get to know about Differentiation under the sign more in-depth in brilliant.org/wiki/differentiate-through-the-integral You will notice that the integral is not fit to apply that method, as you will have to set a function, differentiate it, and anti-differentiate it again which is really hard in here. Too bad you think trig substitution annoying, but I guess that is the fastest, and easiest way :) $\endgroup$ – user547075 May 26 at 1:13
  • $\begingroup$ Usually $/alpha$ is not a square of a number but it can be. $\endgroup$ – Victor S. May 26 at 1:13
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Let $I= \int\sqrt{\alpha-x^2}dx$ $\alpha>0$

Let $u = \sqrt{\alpha-x^2}$ and $dv=dx$

$du = \frac{-x}{\sqrt{\alpha-x^2}}$ and $v = x$

So, $I = \int udv = uv - \int vdu +c$

(Integration by parts)

$I = x\sqrt{\alpha-x^2}-\int{x\frac{-x}{\sqrt{\alpha-x^2}}dx}+c$

$I = x\sqrt{\alpha-x^2}-\int{\frac{\alpha-x^2-\alpha}{\sqrt{\alpha-x^2}}dx}+c = x\sqrt{\alpha-x^2}-\int{\frac{\alpha-x^2}{\sqrt{\alpha-x^2}}dx}+\int\frac{\alpha}{\sqrt{\alpha-x^2}}+c$

$I = x\sqrt{\alpha-x^2} - \int\sqrt{\alpha-x^2}dx+\alpha sin^{-1}\frac{x}{\sqrt{\alpha}}+c$

$I = x\sqrt{\alpha-x^2} - I+\alpha sin^{-1}\frac{x}{\sqrt{\alpha}}+c$

$$I =\frac{x}{2} \sqrt{\alpha-x^2} + \frac{\alpha}{2}sin^{-1}\frac{x}{\sqrt{\alpha}}+C$$

$C = c/2$

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you can use $$x=\sqrt{\alpha }\frac{1-t}{1+t}$$ to get $$-4\alpha \int{\frac{\sqrt{t}}{{{\left( 1+t \right)}^{3}}}dt}$$

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