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I'm trying to compute the antiderivative $$\int \frac{y^2}{\sqrt{r^2 - y^2}} \, dy.$$ It is proving fairly tricky (for me).

Here is Wolfram|Alpha's solution: $$\int \frac{y^2}{\sqrt{r^2 - y^2}} \, dy = \frac{1}{2} \left( r^2 \arctan \left( \frac{y}{\sqrt{r^2 - y^2}} \right) - y\sqrt{r^2 - y^2} \right).$$

The provided step-by-step explanation is:

  • Substitute $y = r \sin u$ and $dy = r \cos u \, du$. Then $\sqrt{r^2 - y^2} = \sqrt{r^2 - r^2 \sin^2 u} = r \cos u$ and $u = \arcsin(y/r)$.
  • The integral becomes $$\int r^2 \sin^2 u \, du = r^2 \int \sin^2 u \, du.$$
  • From here, the antiderivative is pretty commonly known; it can be derived with the double-angle formula for cosine. This becomes $$\frac{1}{2} r^2 u - \frac{1}{4} r^2 \sin(2u).$$
  • Use $\cos^2 u = 1 - \sin^2 u$ and $\sin(2u) = 2\sin u\cos u$ to express as $$\frac{1}{2} r^2 u - \frac{1}{4} r^2 \sin(u) \sqrt{1 - \sin^2 u}.$$
  • Back-substitute $u = \arcsin(y/r)$ to get $$\frac{1}{2} r^2 \arcsin(y/r) - \frac{1}{2} r y \sqrt{1 - \frac{y^2}{r^2}}.$$
  • For positive reals, this is equivalent to $$\frac{1}{2} \left(r^2 \arctan{\frac{y}{\sqrt{r^2 - y^2}}} - y\sqrt{r^2 - y^2}\right).$$

In my opinion, this is a mess, both in derivation and result, and it's not the first time I've seen Wolfram|Alpha overcomplicate an integral.

However, I can't figure out how to solve this myself. I tried $u$-substitution, but the $\Phi(x)$ that might have worked was not injective on the domain so that didn't work

So, is there either or both of

  • a cleaner solution to this antiderivative? or
  • a cleaner method of derivation?

…it would be really nice if it didn't have any trigonometric substitution…

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    $\begingroup$ Pretty standard sub and nice-and-clean solution. Not sure what exactly is complicated there. Would've done exactly the same way. $\endgroup$ – Kaster Jun 11 '14 at 3:41
  • $\begingroup$ That trigonometric substitution is extremely standard when you see things involving $\sqrt{a^2 - x^2}$. $\endgroup$ – user61527 Jun 11 '14 at 3:44
  • $\begingroup$ @Kaster so you don't think there's a cleaner form for the final antiderivative? This is from a textbook, and previous integrations in this section have been along the lines of $\iint_R xy \, dx \, dy$ over a rectangle and other simple integrals. (The chapter is on vector calculus, not integrals specifically; integrals are just a part of it.) $\endgroup$ – wchargin Jun 11 '14 at 3:44
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    $\begingroup$ When the transform function is not injective, split the domain into pieces on each of which it is injective. $\endgroup$ – Eric Towers Jun 11 '14 at 3:47
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I doubt that you can entirely avoid a trig substitution (unless you want to try hyperbolics). However you can go some of the way using integration by parts: taking $$u=y\ ,\quad \frac{dv}{dy}=\frac{y}{\sqrt{r^2-y^2}}$$ leads to $$\int\frac{y^2}{\sqrt{r^2-y^2}}=-y\sqrt{r^2-y^2}+\int \sqrt{r^2-y^2}\,dy$$ and the integral on the right hand side, though still best done with a trig substitution, is a bit easier than the one you started with.

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In my opinion, the result is not a mess and I shall try to explain. Let us start with $$x=\int \frac{y^2}{\sqrt{r^2 - y^2}} \, dy$$ and make the change of variable $$y=r~\sin(u)$$ So, we arrive to $$x=\frac{r^2}{4} \Big(2 u - \sin(2u)\Big)$$ This gives a parametric formulation of $x$ and $y$ as functions of $u$ and, for many applications, it could be the end of the problem.

The problem is that, if you want an explicit expression of $x$ as a function of $y$, you need to go through the next steps and this is where rather complex expressions appear since $x$ contains $u,\sin(u),\cos(u)$.

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To a notable extent, whether something is "complicated" is subjective. The techniques of integration used here are standard, well-known, and not particularly onerous for a first year undergraduate calculus course, or even a high school calculus course.

Moreover, the form of the antiderivative itself strongly suggests that some kind of trigonometric substitution is not only desirable, but inevitable.

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Or do a trick: $\dfrac{y^2}{\sqrt{r^2-y^2}} = -\sqrt{r^2-y^2} + \dfrac{r^2}{\sqrt{r^2-y^2}}$. At this point, we can take $y = r\cdot \sin\theta$.

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