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…