# Integral of $\sqrt{a-x^2}$

I know how to integrate stuff like $\sqrt{a-x^2}$, though I haven't done this in a while. So I wanted to check the answer after calculating. My calculations go like this ($\theta=sin^{-1}(\frac{x}{\sqrt{a}})$): $$\int_{A}^{B}{\sqrt{a-x^2}}dx=\int_{A}^{B}{\sqrt{a-x^2}}\frac{1}{\frac{1}{\sqrt{a}}\frac{1}{\sqrt{1-\frac{x^2}{a}}}}d(sin^{-1}(\frac{x}{\sqrt{a}}))=\int_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}{(a-x^2)d\theta}=\int_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}{(a-((sin\theta)\sqrt{a})^2)d\theta}=[a\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}-a\int_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}{(\frac{1}{2}-\frac{1}{2}cos2\theta)d\theta}=[a\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}-[a\frac{1}{2}\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}+\frac{1}{2}a\int_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}{cos2\theta\frac{1}{2}d(2\theta)}=[\frac{1}{2}a\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}+\frac{1}{2}a[\frac{1}{2}sin2\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}=[\frac{1}{2}a\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}+[\frac{1}{2}a sin\theta cos\theta]_{sin^{-1}(\frac{A}{\sqrt{a}})}^{sin^{-1}(\frac{B}{\sqrt{a}})}=[\frac{1}{2}asin^{-1}(\frac{x}{\sqrt{a}})+\frac{1}{2}x\sqrt{a-x^2}]_{A}^{B}$$

However this doesn't seem to be right after I checked it multiple times with Wolfram Integrator and my calculator. Where did I go wrong?

• You went wrong trying to do too many substitutions at once. First just try $x=v\sqrt{a}$. The do $v=\sin\theta$, etc. Doing them all at once is a nightmare. Commented Aug 25, 2013 at 13:43
• I don't know where, if anywhere, you went wrong, but I would have approached the problem by substituting $x=\sqrt a\sin u$, $dx=\sqrt a\cos u\,du$. Commented Aug 25, 2013 at 13:44

## 1 Answer

So you've got the integral:

$$\int_A^B \sqrt{a - x^2} \,dx = \int_A^B \sqrt{(\sqrt a)^2 - x^2}\,dx$$

Integrals in the form $\sqrt{a^2 - x^2}$ correspond nicely to using the substitution $x = a \sin \theta \implies \,dx = a \cos \theta d\theta$.

In your case, we have $\sqrt{(\sqrt{a})^2 - x^2}$, so our nice substitution would be $x = \sqrt a \sin\theta$. Such a substitution from the start greatly simplifies the calculation.

For a refresher on using trigonometric substitution, see the corresponding Wikipedia entry, which starts off integrands containing $a^2 - x^2$, and the use of the substitution $x = a \sin \theta$. There are a few other handy substitutions to use in some other choice situations posted in that same entry.

• Very helpful advice +1 Commented Aug 25, 2013 at 14:16
• @amWhy $^+_+$$\ddot\smile^+_+$ Commented Aug 25, 2013 at 14:44
• @amWhy If you look at what I did ($\theta=sin^{-1}(\frac{x}{\sqrt{a}})$), its exactly the same to the substitution $x=\sqrt{a}sin\theta$. Commented Aug 25, 2013 at 14:47
• @amWhy Alright I did the calculation with the substitution $x=\sqrt{a}sin\theta$. I obtain the same results, in conflict with my checking. Can anyone else confirm that using $x=\sqrt{a}sin\theta$ gives results equal to my original result? Commented Aug 25, 2013 at 15:08
• Your work is correct, a bit more difficult than need be. Your result is correct. Commented Aug 25, 2013 at 16:01