Integration practice of $\int \frac{\sqrt{25-y^2}}{y}dy$

I need to solve $\int \frac{\sqrt{25-y^2}}{y}dy$.

I originally thought IBP, but that led to a very large and confusing algebra problem.

Then I started to look at the $\sqrt{25-y^2}$ and started to consider doing a $u$-substituion, and letting $u=\sin\theta$ and $du=\cos\theta d\theta$. Once I started playing a bit, the intuition was lost.

Any and all help is appreciated.

Solution 1:

Trigonometric substitutions are not necessary. Let $u^2 = 25 - y^2$, so that $2u \, du = -2y \, dy$. Then $$\begin{align*} \int \frac{\sqrt{25-y^2}}{y} \, dy &= - \int \frac{-y \sqrt{25-y^2}}{y^2} \, dy \\ &= - \int \frac{u \sqrt{u^2}}{25-u^2} \, du \\ &= \int \frac{u^2 - 25 + 25}{u^2 - 25} \, du \\ &= u + 25 \int \frac{du}{u^2 - 25} \\ &= u + \frac{5}{2} \int \left(\frac{1}{u-5} - \frac{1}{u+5} \,\right) du\end{align*}$$ and the rest is obvious.

Solution 2:

Let $y=5 \sin u$, then you have

\begin{eqnarray}{c} \int \frac{\sqrt{25-y^2}}{y}{\rm d}y &=& \int \frac{\sqrt{25-25\sin^2u}}{5\sin u}5\cos u\ {\rm d}{u} \\& =& \int \frac{\cos u}{\sin u}5\cos u\ {\rm d}{u} \\& =& 5\int \frac{\cos^2 u}{\sin u}\ {\rm d}{u} \\& =& 5\int \frac{1-\sin^2 u}{\sin u}\ {\rm d}{u} \\& =& 5\int \frac{1}{\sin u}-\sin u\ {\rm d}{u} \\& =& 5\int \csc u-\sin u\ {\rm d}{u} \\& =& \ -5\ln \left |\csc u + \cot u\right | - 5\cos u + C. \end{eqnarray} Now substitute $u$ with $\arcsin y$.