1. **Problem statement:** Given a function $f$ with continuous second derivative on $\mathbb{R}$, and conditions $f'(0)=-1$, $f(1)=0$, and $f'(1)=0$, prove that $$\int_0^1 f(x)\cosh(x)\,dx = -1 + \int_0^1 f''(x)\cosh(x)\,dx.$$\n\n2. **Recall integration by parts formula:** $$\int u\,dv = uv - \int v\,du.$$\n\n3. **Choose:** - $u = f(x)$ so that $du = f'(x)\,dx$ - $dv = \cosh(x)\,dx$ so that $v = \sinh(x)$ \n4. **Apply integration by parts:** $$\int_0^1 f(x)\cosh(x)\,dx = \left[f(x)\sinh(x)\right]_0^1 - \int_0^1 f'(x)\sinh(x)\,dx.$$\n\n5. **Evaluate boundary term:** $$f(1)\sinh(1) - f(0)\sinh(0) = 0 \cdot \sinh(1) - f(0) \cdot 0 = 0.$$\n\n6. **Now consider the integral:** $$\int_0^1 f'(x)\sinh(x)\,dx.$$\nUse integration by parts again with: - $u = f'(x)$ so $du = f''(x)\,dx$ - $dv = \sinh(x)\,dx$ so $v = \cosh(x)$ \n7. **Apply integration by parts second time:** $$\int_0^1 f'(x)\sinh(x)\,dx = \left[f'(x)\cosh(x)\right]_0^1 - \int_0^1 f''(x)\cosh(x)\,dx.$$\n\n8. **Evaluate boundary term:** $$f'(1)\cosh(1) - f'(0)\cosh(0) = 0 \cdot \cosh(1) - (-1) \cdot 1 = 1.$$\n\n9. **Substitute back:** $$\int_0^1 f(x)\cosh(x)\,dx = 0 - \left(1 - \int_0^1 f''(x)\cosh(x)\,dx\right) = -1 + \int_0^1 f''(x)\cosh(x)\,dx.$$\n\n**Final answer:** $$\boxed{\int_0^1 f(x)\cosh(x)\,dx = -1 + \int_0^1 f''(x)\cosh(x)\,dx}.$$