1. Problem: Evaluate the integral $$\int_{\pi/4}^{\pi/2} \sin x \frac{d}{dx} + \cos x F(X) \, dx$$. 2. Interpretation: The integral as written is ambiguous. It seems to involve $$\sin x \frac{d}{dx}$$ (a differential operator) plus $$\cos x F(X)$$ inside an integral with respect to $$x$$. 3. Usually, $$\frac{d}{dx}$$ denotes differentiation wrt $$x$$, but it cannot stand alone inside an integral without being applied to a function. 4. Assuming the problem intends to evaluate $$\int_{\pi/4}^{\pi/2} \sin x \frac{d}{dx}[F(x)] + \cos x F(x) \, dx$$, where $$F(x)$$ is a function of $$x$$. 5. Rewrite the integral as $$\int_{\pi/4}^{\pi/2} [\sin x F'(x) + \cos x F(x)] \, dx$$ where $$F'(x) = \frac{d}{dx}F(x)$$. 6. Notice that $$\frac{d}{dx}[\sin x F(x)] = \cos x F(x) + \sin x F'(x)$$ by product rule. 7. Therefore the integral becomes $$\int_{\pi/4}^{\pi/2} \frac{d}{dx}[\sin x F(x)] \, dx$$. 8. By the Fundamental Theorem of Calculus, this evaluates to $$\sin x F(x) \Big|_{\pi/4}^{\pi/2} = \sin\frac{\pi}{2} F\left(\frac{\pi}{2}\right) - \sin\frac{\pi}{4} F\left(\frac{\pi}{4}\right)$$. 9. Simplify the sine values: $$\sin\frac{\pi}{2} = 1$$ and $$\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$. 10. Final answer: $$F\left(\frac{\pi}{2}\right) - \frac{\sqrt{2}}{2} F\left(\frac{\pi}{4}\right)$$. Note: The result depends on the function $$F(x)$$ at the integration limits.