1. The problem is to find the primitive (antiderivative) of the function $f(x) = \arctan(x) - \sin(x)$ on the interval $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$. 2. The primitive of a function $f(x)$ is a function $F(x)$ such that $F'(x) = f(x)$. 3. We find the primitive by integrating each term separately: $$\int \arctan(x) \, dx - \int \sin(x) \, dx$$ 4. For $\int \arctan(x) \, dx$, use integration by parts: Let $u = \arctan(x)$, so $du = \frac{1}{1+x^2} dx$. Let $dv = dx$, so $v = x$. Then: $$\int \arctan(x) \, dx = x \arctan(x) - \int \frac{x}{1+x^2} dx$$ 5. Simplify the integral: $$\int \frac{x}{1+x^2} dx$$ Use substitution $w = 1 + x^2$, so $dw = 2x dx$, thus $x dx = \frac{dw}{2}$. Therefore: $$\int \frac{x}{1+x^2} dx = \int \frac{1}{w} \cdot \frac{dw}{2} = \frac{1}{2} \int \frac{1}{w} dw = \frac{1}{2} \ln|w| + C = \frac{1}{2} \ln(1+x^2) + C$$ 6. So the primitive of $\arctan(x)$ is: $$x \arctan(x) - \frac{1}{2} \ln(1+x^2) + C$$ 7. The primitive of $-\sin(x)$ is: $$- \int \sin(x) dx = -(-\cos(x)) = \cos(x) + C$$ 8. Combine both results: $$F(x) = x \arctan(x) - \frac{1}{2} \ln(1+x^2) + \cos(x) + C$$ 9. To find the definite integral (primitive evaluated on the interval), compute: $$F\left(\frac{\pi}{2}\right) - F\left(\frac{\pi}{4}\right)$$ 10. Substitute values: $$F\left(\frac{\pi}{2}\right) = \frac{\pi}{2} \arctan\left(\frac{\pi}{2}\right) - \frac{1}{2} \ln\left(1+\left(\frac{\pi}{2}\right)^2\right) + \cos\left(\frac{\pi}{2}\right)$$ $$F\left(\frac{\pi}{4}\right) = \frac{\pi}{4} \arctan\left(\frac{\pi}{4}\right) - \frac{1}{2} \ln\left(1+\left(\frac{\pi}{4}\right)^2\right) + \cos\left(\frac{\pi}{4}\right)$$ 11. Note that $\cos\left(\frac{\pi}{2}\right) = 0$ and $\arctan\left(\frac{\pi}{2}\right)$ is approximately $1.0038848$ radians. 12. The final answer is: $$\left(\frac{\pi}{2} \arctan\left(\frac{\pi}{2}\right) - \frac{1}{2} \ln\left(1+\left(\frac{\pi}{2}\right)^2\right) + 0\right) - \left(\frac{\pi}{4} \arctan\left(\frac{\pi}{4}\right) - \frac{1}{2} \ln\left(1+\left(\frac{\pi}{4}\right)^2\right) + \cos\left(\frac{\pi}{4}\right)\right)$$