1. The problem is to evaluate the integral $$\int_0^{\frac{\pi}{2}} \cos^3 x \, dx$$. 2. We use the reduction formula or rewrite the power of cosine using trigonometric identities. Recall that $$\cos^3 x = \cos x \cdot \cos^2 x = \cos x (1 - \sin^2 x)$$. 3. Substitute this into the integral: $$\int_0^{\frac{\pi}{2}} \cos^3 x \, dx = \int_0^{\frac{\pi}{2}} \cos x (1 - \sin^2 x) \, dx = \int_0^{\frac{\pi}{2}} \cos x \, dx - \int_0^{\frac{\pi}{2}} \cos x \sin^2 x \, dx$$. 4. Let us evaluate each integral separately. 5. First integral: $$\int_0^{\frac{\pi}{2}} \cos x \, dx = \sin x \Big|_0^{\frac{\pi}{2}} = \sin \frac{\pi}{2} - \sin 0 = 1 - 0 = 1$$. 6. For the second integral, use substitution: let $$u = \sin x$$, then $$du = \cos x \, dx$$. 7. Change the limits accordingly: when $$x=0$$, $$u=\sin 0=0$$; when $$x=\frac{\pi}{2}$$, $$u=\sin \frac{\pi}{2}=1$$. 8. The second integral becomes: $$\int_0^{\frac{\pi}{2}} \cos x \sin^2 x \, dx = \int_0^1 u^2 \, du = \frac{u^3}{3} \Big|_0^1 = \frac{1}{3} - 0 = \frac{1}{3}$$. 9. Combine the results: $$\int_0^{\frac{\pi}{2}} \cos^3 x \, dx = 1 - \frac{1}{3} = \frac{2}{3}$$. 10. Therefore, the value of the integral is $$\boxed{\frac{2}{3}}$$.