The integral in this exercise converges. Evaluate the integral without using a table. \( \int_{0}^{5} \frac{\mathrm{dx}}{\sqrt{25-\mathrm{x}^{2}}} \)

To evaluate the integral \( \int_{0}^{5} \frac{\mathrm{dx}}{\sqrt{25-\mathrm{x}^{2}}} \), we can use the trigonometric substitution \( x = 5 \sin(\theta) \). This leads to \( dx = 5 \cos(\theta) d\theta \) and changes the limits of integration. When \( x = 0 \), \( \theta = 0 \), and when \( x = 5 \), \( \theta = \frac{\pi}{2} \). Substituting in, we get: \[ \int_{0}^{\frac{\pi}{2}} \frac{5 \cos(\theta) \, d\theta}{\sqrt{25 - 25 \sin^2(\theta)}} = \int_{0}^{\frac{\pi}{2}} \frac{5 \cos(\theta) \, d\theta}{\sqrt{25(1 - \sin^2(\theta))}} = \int_{0}^{\frac{\pi}{2}} \frac{5 \cos(\theta) \, d\theta}{5 \cos(\theta)} = \int_{0}^{\frac{\pi}{2}} d\theta \] This integral simplifies to: \[ \int_{0}^{\frac{\pi}{2}} d\theta = \left[ \theta \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \] Thus, the value of the integral is: \[ \int_{0}^{5} \frac{\mathrm{dx}}{\sqrt{25-\mathrm{x}^{2}}} = \frac{\pi}{2} \]