Evaluate the integral. \( \int_{0}^{6} \frac{d t}{\sqrt{36+t^{2}}} \)

To evaluate the integral \( \int_{0}^{6} \frac{dt}{\sqrt{36+t^{2}}} \), we can use the trigonometric substitution \( t = 6 \tan(\theta) \). This gives \( dt = 6 \sec^2(\theta) d\theta \) and the limits change from \( t=0 \) to \( t=6 \) which corresponds to \( \theta=0 \) to \( \theta=\frac{\pi}{4} \). Substituting these into the integral, we have: \[ \int_{0}^{\frac{\pi}{4}} \frac{6 \sec^2(\theta) d\theta}{\sqrt{36 + 36\tan^2(\theta)}} \] This simplifies to: \[ \int_{0}^{\frac{\pi}{4}} \frac{6 \sec^2(\theta)}{\sqrt{36(1 + \tan^2(\theta))}} d\theta = \int_{0}^{\frac{\pi}{4}} \frac{6 \sec^2(\theta)}{\sqrt{36 \sec^2(\theta)}} d\theta \] The square root simplifies: \[ = \int_{0}^{\frac{\pi}{4}} \frac{6 \sec^2(\theta)}{6 \sec(\theta)} d\theta = \int_{0}^{\frac{\pi}{4}} \sec(\theta) d\theta \] The integral of \( \sec(\theta) \) is: \[ \int \sec(\theta) d\theta = \ln | \sec(\theta) + \tan(\theta) | + C \] Thus, we evaluate: \[ \left[ \ln | \sec(\theta) + \tan(\theta) | \right]_{0}^{\frac{\pi}{4}} \] At \( \theta = \frac{\pi}{4} \): \[ \sec\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad \tan\left(\frac{\pi}{4}\right) = 1 \quad \Rightarrow \quad \ln | \sqrt{2} + 1 | \] At \( \theta = 0 \): \[ \sec(0) = 1, \quad \tan(0) = 0 \quad \Rightarrow \quad \ln | 1 + 0 | = \ln(1) = 0 \] Putting it together, we have: \[ \ln | \sqrt{2} + 1 | - 0 = \ln | \sqrt{2} + 1 | \] Thus the value of the integral is: \[ \int_{0}^{6} \frac{dt}{\sqrt{36+t^{2}}} = \ln(\sqrt{2} + 1) \]