1. The problem is to find which substitution transforms the integral $$\int \frac{\sqrt{y^2 - 81}}{y} \, dy$$ with the condition $y > 9$ into an integral that can be evaluated directly in terms of $\theta$. 2. Notice the expression under the square root is $y^2 - 81$, which can be written as $y^2 - 9^2$. 3. For integrals involving $\sqrt{y^2 - a^2}$, the standard trigonometric substitution is $y = a \sec \theta$ because: $$\sqrt{y^2 - a^2} = \sqrt{a^2 \sec^2 \theta - a^2} = a \sqrt{\sec^2 \theta - 1} = a \tan \theta$$ 4. Since $a = 9$, the substitution is: $$y = 9 \sec \theta$$ 5. This substitution simplifies the integral to a form involving $\tan \theta$ and $\sec \theta$, which can be integrated using standard trigonometric integrals. 6. The other options, $y = 9 \sin \theta$ or $y = 9 \tan \theta$, are used for $\sqrt{a^2 - y^2}$ and $\sqrt{a^2 + y^2}$ respectively, which do not match the form here. Final answer: C. $y = 9 \sec \theta$