1. **State the problem:** We want to solve the integral $$\int \frac{\sqrt{a^2 - y^2}}{y} \, dy$$ where $a$ is a constant. 2. **Recall the formula and substitution:** This integral involves a square root of the form $\sqrt{a^2 - y^2}$, which suggests a trigonometric substitution. We use the substitution: $$y = a \sin \theta$$ which implies $$dy = a \cos \theta \, d\theta$$ 3. **Rewrite the integral in terms of $\theta$:** Substitute $y$ and $dy$: $$\sqrt{a^2 - y^2} = \sqrt{a^2 - a^2 \sin^2 \theta} = \sqrt{a^2 (1 - \sin^2 \theta)} = a \cos \theta$$ The integral becomes: $$\int \frac{a \cos \theta}{a \sin \theta} \cdot a \cos \theta \, d\theta = \int \frac{a \cos \theta}{a \sin \theta} \cdot a \cos \theta \, d\theta = \int \frac{a^2 \cos^2 \theta}{a \sin \theta} \, d\theta = \int \frac{a \cos^2 \theta}{\sin \theta} \, d\theta$$ 4. **Simplify the integral:** $$\int \frac{a \cos^2 \theta}{\sin \theta} \, d\theta = a \int \frac{\cos^2 \theta}{\sin \theta} \, d\theta$$ Use the identity $\cos^2 \theta = 1 - \sin^2 \theta$: $$a \int \frac{1 - \sin^2 \theta}{\sin \theta} \, d\theta = a \int \left( \frac{1}{\sin \theta} - \sin \theta \right) d\theta = a \int \csc \theta \, d\theta - a \int \sin \theta \, d\theta$$ 5. **Integrate each term:** - Integral of $\csc \theta$ is: $$\int \csc \theta \, d\theta = \ln \left| \tan \frac{\theta}{2} \right| + C$$ - Integral of $\sin \theta$ is: $$\int \sin \theta \, d\theta = -\cos \theta + C$$ So, $$a \int \csc \theta \, d\theta - a \int \sin \theta \, d\theta = a \ln \left| \tan \frac{\theta}{2} \right| + a \cos \theta + C$$ 6. **Back-substitute to $y$:** Recall $y = a \sin \theta$, so $$\sin \theta = \frac{y}{a}$$ and $$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{y^2}{a^2}} = \frac{\sqrt{a^2 - y^2}}{a}$$ Also, $$\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}$$ Substitute $\cos \theta$: $$\tan \frac{\theta}{2} = \sqrt{\frac{1 - \frac{\sqrt{a^2 - y^2}}{a}}{1 + \frac{\sqrt{a^2 - y^2}}{a}}} = \sqrt{\frac{a - \sqrt{a^2 - y^2}}{a + \sqrt{a^2 - y^2}}}$$ 7. **Final answer:** $$\int \frac{\sqrt{a^2 - y^2}}{y} \, dy = a \ln \left| \sqrt{\frac{a - \sqrt{a^2 - y^2}}{a + \sqrt{a^2 - y^2}}} \right| + \sqrt{a^2 - y^2} + C$$ This completes the solution.