I have the following integral:

$$ I =\int x^2\sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) dx$$

Where $a$ and $b$ are non-zero positive integers, and $x<a$. I have started by integration by parts:

$$ I = uv - \int vu'$$

$$u = \sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right )$$

$$u' = \left [ \sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) \right ]'$$

We substitute the inside content of the arcsin by $z=\frac{\sqrt{a^2-x^2}}{b}$:

$$u' = \left [ \sin^{-1}(z) \right ]' = \frac{1}{\sqrt{1-z^2}} \cdot z'$$

$$z' = - \frac{x}{b} (a^2-x^2)^{- \frac{1}{2}}$$

Hence we obtain:

$$u' = -\frac{x}{\sqrt{(a^2-x^2)(b^2-a^2+x^2)}} $$

On the other hand of the integration by parts we have $v'$:

$$v' = x^2$$

$$v = \frac{1}{3}x^3$$

Hence we end up with:

$$I = \frac{1}{3}x^3\sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) + \frac{1}{3}\int \frac{x^4}{\sqrt{(a^2-x^2)(b^2-a^2+x^2)}} $$

I cant really figure out how to solve this final integral. Any help would be appreciated

  • 2
    $\begingroup$ If you type \mathrm{sin} instead of \sin then you'll see $2\mathrm{sin}x$ instead of $2\sin x.$ I edited accordingly. $\endgroup$ Mar 22, 2021 at 20:01
  • 2
    $\begingroup$ Looks like an elliptic integral, however you slice it. $\endgroup$ Mar 22, 2021 at 20:24
  • $\begingroup$ Thanks @TedShifrin . It ended up being an elliptic integral as you said :( It was eitherway semi-useful for my context, given that MATLab can plot it. $\endgroup$ Mar 29, 2021 at 16:25

1 Answer 1


Make the substitution

$$u = \frac{\sqrt{a^2-x^2}}b.$$

Then $x = \sqrt{a^2 - u^2 b^2},$ and so $d x = - \frac{b^2 du}{2 \sqrt{a^2-u^2}},$ and so the integral becomes

$$\frac{b^2}2\int \sqrt{a^2 - u^2 b^2} \arcsin(u) du= \frac{b^2 a}{2} \int \sqrt{1- (\frac{ub}a)^2} \arcsin u du$$ Now integrate by parts.

  • $\begingroup$ Am I missing something? Have you actually tried to do this? $\endgroup$ Mar 22, 2021 at 20:18
  • $\begingroup$ Don't you mean $$dx=-\frac{b^2 u du}{\sqrt{a^2-u^2b^2}}?$$ I think you missed a few bits. $\endgroup$ Mar 22, 2021 at 20:41
  • $\begingroup$ Once it's fixed up, it's going to be the same elliptic integral as before ... $\endgroup$ Mar 22, 2021 at 20:43
  • $\begingroup$ @TedShifrin Yes, of course it is. What's your point? $\endgroup$
    – Igor Rivin
    Mar 23, 2021 at 0:06
  • 1
    $\begingroup$ So, first, you have an elementary error, and then you suggest that it'll be easy to integrate by parts. How is this any help to the OP? My point is that this is of negative constructive value. $\endgroup$ Mar 23, 2021 at 0:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .