1. **State the problem:** We need to evaluate the integral $$\int_0^{\frac{5}{\sqrt{2}}} \frac{dx}{\sqrt{25 - x^2}}$$ and determine which substitution transforms it into an integral in terms of $\theta$ that can be evaluated directly. 2. **Recognize the integral form:** The integral is of the form $$\int \frac{dx}{\sqrt{a^2 - x^2}}$$ where $a=5$. 3. **Recall the standard trigonometric substitution:** For integrals involving $$\sqrt{a^2 - x^2}$$, the substitution $$x = a \sin \theta$$ is used because: - It simplifies $$\sqrt{a^2 - x^2}$$ to $$a \cos \theta$$. - The differential $$dx = a \cos \theta d\theta$$. 4. **Apply the substitution:** Using $$x = 5 \sin \theta$$, - $$dx = 5 \cos \theta d\theta$$ - $$\sqrt{25 - x^2} = \sqrt{25 - 25 \sin^2 \theta} = 5 \cos \theta$$ 5. **Rewrite the integral:** $$\int \frac{dx}{\sqrt{25 - x^2}} = \int \frac{5 \cos \theta d\theta}{5 \cos \theta} = \int d\theta$$ which is straightforward to integrate. 6. **Determine the new limits:** When $$x=0$$, $$0 = 5 \sin \theta \Rightarrow \theta = 0$$. When $$x= \frac{5}{\sqrt{2}}$$, $$\frac{5}{\sqrt{2}} = 5 \sin \theta \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = \frac{\pi}{4}$$. 7. **Conclusion:** The substitution $$x = 5 \sin \theta$$ (option C) transforms the integral into one that can be evaluated directly in terms of $\theta$. **Final answer:** C. $x = 5 \sin \theta$