1. **Problem statement:** Find the Fourier cosine transform of the piecewise function $$f(x) = \begin{cases} x, & 0 < x < \frac{1}{2} \\ 1 - x, & \frac{1}{2} < x < 1 \\ 0, & x > 1 \end{cases}$$ 2. **Formula for Fourier cosine transform:** The Fourier cosine transform $F_c(\omega)$ of a function $f(x)$ is given by $$F_c(\omega) = \sqrt{\frac{2}{\pi}} \int_0^{\infty} f(x) \cos(\omega x) \, dx$$ 3. **Apply the piecewise definition:** Since $f(x)$ is zero for $x > 1$, the integral limits reduce to $0$ to $1$: $$F_c(\omega) = \sqrt{\frac{2}{\pi}} \left( \int_0^{\frac{1}{2}} x \cos(\omega x) \, dx + \int_{\frac{1}{2}}^1 (1 - x) \cos(\omega x) \, dx \right)$$ 4. **Evaluate the first integral:** Use integration by parts for $$I_1 = \int_0^{\frac{1}{2}} x \cos(\omega x) \, dx$$ Let $u = x$, $dv = \cos(\omega x) dx$, then $du = dx$, $v = \frac{\sin(\omega x)}{\omega}$. So, $$I_1 = \left. x \frac{\sin(\omega x)}{\omega} \right|_0^{\frac{1}{2}} - \int_0^{\frac{1}{2}} \frac{\sin(\omega x)}{\omega} dx = \frac{1}{2} \frac{\sin(\frac{\omega}{2})}{\omega} - \frac{1}{\omega} \int_0^{\frac{1}{2}} \sin(\omega x) dx$$ Evaluate the remaining integral: $$\int_0^{\frac{1}{2}} \sin(\omega x) dx = \left. -\frac{\cos(\omega x)}{\omega} \right|_0^{\frac{1}{2}} = -\frac{\cos(\frac{\omega}{2}) - 1}{\omega}$$ Thus, $$I_1 = \frac{1}{2} \frac{\sin(\frac{\omega}{2})}{\omega} - \frac{1}{\omega} \left(-\frac{\cos(\frac{\omega}{2}) - 1}{\omega} \right) = \frac{1}{2} \frac{\sin(\frac{\omega}{2})}{\omega} + \frac{1 - \cos(\frac{\omega}{2})}{\omega^2}$$ 5. **Evaluate the second integral:** $$I_2 = \int_{\frac{1}{2}}^1 (1 - x) \cos(\omega x) \, dx = \int_{\frac{1}{2}}^1 \cos(\omega x) dx - \int_{\frac{1}{2}}^1 x \cos(\omega x) dx$$ First integral: $$\int_{\frac{1}{2}}^1 \cos(\omega x) dx = \left. \frac{\sin(\omega x)}{\omega} \right|_{\frac{1}{2}}^1 = \frac{\sin(\omega) - \sin(\frac{\omega}{2})}{\omega}$$ Second integral (by parts): Let $u = x$, $dv = \cos(\omega x) dx$, so $du = dx$, $v = \frac{\sin(\omega x)}{\omega}$. $$\int_{\frac{1}{2}}^1 x \cos(\omega x) dx = \left. x \frac{\sin(\omega x)}{\omega} \right|_{\frac{1}{2}}^1 - \int_{\frac{1}{2}}^1 \frac{\sin(\omega x)}{\omega} dx = \frac{\sin(\omega)}{\omega} - \frac{1}{2} \frac{\sin(\frac{\omega}{2})}{\omega} - \frac{1}{\omega} \int_{\frac{1}{2}}^1 \sin(\omega x) dx$$ Evaluate the remaining integral: $$\int_{\frac{1}{2}}^1 \sin(\omega x) dx = \left. -\frac{\cos(\omega x)}{\omega} \right|_{\frac{1}{2}}^1 = -\frac{\cos(\omega) - \cos(\frac{\omega}{2})}{\omega}$$ So, $$\int_{\frac{1}{2}}^1 x \cos(\omega x) dx = \frac{\sin(\omega)}{\omega} - \frac{1}{2} \frac{\sin(\frac{\omega}{2})}{\omega} + \frac{\cos(\omega) - \cos(\frac{\omega}{2})}{\omega^2}$$ Therefore, $$I_2 = \frac{\sin(\omega) - \sin(\frac{\omega}{2})}{\omega} - \left( \frac{\sin(\omega)}{\omega} - \frac{1}{2} \frac{\sin(\frac{\omega}{2})}{\omega} + \frac{\cos(\omega) - \cos(\frac{\omega}{2})}{\omega^2} \right) = - \frac{\sin(\frac{\omega}{2})}{\omega} + \frac{1}{2} \frac{\sin(\frac{\omega}{2})}{\omega} - \frac{\cos(\omega) - \cos(\frac{\omega}{2})}{\omega^2} = - \frac{1}{2} \frac{\sin(\frac{\omega}{2})}{\omega} - \frac{\cos(\omega) - \cos(\frac{\omega}{2})}{\omega^2}$$ 6. **Combine integrals:** $$F_c(\omega) = \sqrt{\frac{2}{\pi}} (I_1 + I_2) = \sqrt{\frac{2}{\pi}} \left( \frac{1}{2} \frac{\sin(\frac{\omega}{2})}{\omega} + \frac{1 - \cos(\frac{\omega}{2})}{\omega^2} - \frac{1}{2} \frac{\sin(\frac{\omega}{2})}{\omega} - \frac{\cos(\omega) - \cos(\frac{\omega}{2})}{\omega^2} \right)$$ Simplify: $$F_c(\omega) = \sqrt{\frac{2}{\pi}} \left( \frac{1 - \cos(\frac{\omega}{2})}{\omega^2} - \frac{\cos(\omega) - \cos(\frac{\omega}{2})}{\omega^2} \right) = \sqrt{\frac{2}{\pi}} \frac{1 - \cos(\frac{\omega}{2}) - \cos(\omega) + \cos(\frac{\omega}{2})}{\omega^2} = \sqrt{\frac{2}{\pi}} \frac{1 - \cos(\omega)}{\omega^2}$$ 7. **Final answer:** $$\boxed{F_c(\omega) = \sqrt{\frac{2}{\pi}} \frac{1 - \cos(\omega)}{\omega^2}}$$ This is the Fourier cosine transform of the given piecewise function.