1. **State the problem:** Evaluate the integral $$\int_0^{\frac{5}{\sqrt{2}}} \frac{dx}{\sqrt{25 - x^2}}.$$\n\n2. **Identify the substitution:** The integral has the form $$\int \frac{dx}{\sqrt{a^2 - x^2}}$$ where $$a=5$$. A common trigonometric substitution for this form is $$x = a \sin \theta$$ because $$1 - \sin^2 \theta = \cos^2 \theta$$ simplifies the square root.\n\n3. **Choose substitution:** From the options, the correct substitution is \textbf{C.} $$x = 5 \sin \theta$$.\n\n4. **Compute differential:** Given $$x = 5 \sin \theta$$, then $$dx = 5 \cos \theta \, d\theta$$.\n\n5. **Change limits:** When $$x=0$$, $$0 = 5 \sin \theta \Rightarrow \sin \theta = 0 \Rightarrow \theta = 0$$.\nWhen $$x=\frac{5}{\sqrt{2}}$$, $$\frac{5}{\sqrt{2}} = 5 \sin \theta \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = \frac{\pi}{4}$$.\n\n6. **Rewrite the integral:**\n$$\int_0^{\frac{5}{\sqrt{2}}} \frac{dx}{\sqrt{25 - x^2}} = \int_0^{\frac{\pi}{4}} \frac{5 \cos \theta \, d\theta}{\sqrt{25 - 25 \sin^2 \theta}} = \int_0^{\frac{\pi}{4}} \frac{5 \cos \theta \, d\theta}{\sqrt{25(1 - \sin^2 \theta)}}.$$\n\n7. **Simplify the square root:**\n$$\sqrt{25(1 - \sin^2 \theta)} = \sqrt{25 \cos^2 \theta} = 5 \cos \theta.$$\n\n8. **Simplify the integrand:**\n$$\frac{5 \cos \theta}{5 \cos \theta} = 1.$$\n\n9. **Integral becomes:**\n$$\int_0^{\frac{\pi}{4}} 1 \, d\theta = \left. \theta \right|_0^{\frac{\pi}{4}} = \frac{\pi}{4} - 0 = \frac{\pi}{4}.$$\n\n**Final answer:** $$\boxed{\frac{\pi}{4}}.$$