1. The problem is to find the differential $dy$ when using the substitution $y = 9 \sec \theta$. 2. Recall the substitution: $$y = 9 \sec \theta$$ 3. To find $dy$, differentiate both sides with respect to $\theta$: $$\frac{dy}{d\theta} = 9 \frac{d}{d\theta}(\sec \theta)$$ 4. The derivative of $\sec \theta$ is $\sec \theta \tan \theta$, so: $$\frac{dy}{d\theta} = 9 \sec \theta \tan \theta$$ 5. Therefore, the differential $dy$ is: $$dy = 9 \sec \theta \tan \theta \, d\theta$$ This expression for $dy$ will be used to substitute in the integral after the trigonometric substitution. Final answer: $$dy = 9 \sec \theta \tan \theta \, d\theta$$