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Is there a simple and nice way to solve $\int\sqrt{a^2-x^2}dx$:

PS:I am not looking for a substitution like $x=a\sin p$,

hrkrshnn
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Tom Lynd
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2 Answers2

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Firstly, I would like to point out that the simplest way to solve this problem is to use trigonometric substitution.

In addition, if you insist on do not using that method, it is surely have some other methods. Let $I$ denote the result of the indefinite integral. We have: \begin{equation} I=\int \frac{a^2-x^2}{\sqrt{a^2-x^2}}dx=a^2\arcsin\left(\frac{x}{a}\right)-\int\frac{x^2}{\sqrt{a^2-x^2}}dx\\[10mm] I=x\sqrt{a^2-x^2}+\int\frac{x^2}{\sqrt{a^2-x^2}}dx \end{equation} The 2nd equation is given by integral by parts. Then add them together, we have \begin{equation} I=\frac{a^2}{2}\arcsin\left(\frac{x}{a}\right)+\frac{x}{2}\sqrt{a^2-x^2} \end{equation} I ignore the constant $C$ in the result.

Git Gud
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Lion
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  • @ Lion, genius,I find it simpler, +1 – Tom Lynd Mar 01 '14 at 10:04
  • @ Lion,just wondering if there's some other genius way to tackle $\int\sqrt{a^2+x^2}dx$:(again no trig substitution) – Tom Lynd Mar 01 '14 at 14:06
  • @TomLynd I reckon that complex method is also a effective way, but it seems more difficult because we will transform the indefinite integral to definite integral then calculate some contour integral which and maybe also use residue theorem. – Lion Mar 02 '14 at 00:15
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Another way to attack such kinds of integrals is using Chebyshev's substitution. Consider the integral: $$ I_{m,p} = \int x^m (\alpha x^n + \beta)^pdx $$

Chebyshev has shown that $I_{m, p}$ is expressed in elementary functions if and only if one of the following cases is satisfied: $$\begin{align*} &1)\ p \in \Bbb Z \\ &2)\ {m+1 \over n} \in \Bbb Z \\ &3)\ {m+1 \over n} + {1\over p} \in \Bbb Z \end{align*} $$

For the first case substitute: $$ x = t^N,\ \text{where}\ N\ \text{is common denominator of fractions}\ m\ \text{and}\ n $$ Second case, substitute: $$ \alpha x^n + \beta = t^M\ \text{where}\ M\ \text{is the denominator of}\ p $$ Third case substitute: $$ \alpha + \beta x^{-n} = t^M\ \text{where}\ M\ \text{is the denominator of}\ p $$

If you check the third case is satisfied, hence the integral exists in elementary functions. $$ (a^2 - x^2)^{1\over 2} = x^m(\alpha x^n + \beta)^p $$ Therefore: $$ p = {1\over 2}\\ m = 0\\ n = 2\\ \alpha = -1\\ \beta = a^2 $$ Let: $$ \alpha + \beta x^{-n} = t^M $$

So: $$ t^2 = {a^2\over x^2} - 1\\ x^2 = {a^2\over t^2 + 1}\\ x^3 = \left({a^2\over t^2 + 1}\right)^{3\over 2}\\ tdt = -{a^2 \over x^3}dx \implies dx = -{t\over a^2}\left({a^2\over t^2 + 1}\right)^{3\over 2}dt $$ The integral becomes: $$\begin{align} I &= -\int \sqrt{a^2 - {a^2\over t^2 + 1}}{t\over a^2}\left({a^2\over t^2 + 1}\right)^{3\over 2}dt \\ &= -\int \sqrt{a^2 t^2 \over t^2 + 1}{t\over a^2}\left({a^2\over t^2 + 1}\right)^{3\over 2}dt\\ &= -\int {|t|t\over a^2}\left({a^2\over t^2 + 1}\right)^2dt \end{align} $$

And this is a regular rational function.

roman
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