The derivative of an integral

Write $e^{x^2/4} F(x) = \int_{\sin(x)}^{\cos(x)}e^{t^2+tx+x^2/4}\,dt = \int_{\sin(x)}^{\cos(x)}e^{(t+x/2)^2}\,dt = \int_{\sin(x)+x/2}^{\cos(x)+x/2} e^{\tau^2}\,d\tau.$

Now take the derivative of both sides, using the chain rule and fundamental theorem of calculus on the right-hand side, $\frac{1}{2}xe^{x^2/4} F(x) + e^{x^2/4} F'(x) = e^{(\cos(x)+x/2)^2}\left(-\sin(x)+\frac{1}{2}\right) - e^{(\sin(x)+x/2)^2}\left(\cos(x) + \frac{1}{2}\right),$ and set $x=0$ : $F'(0) = e^1\left(\frac{1}{2}\right) - e^0\left(1+\frac{1}{2}\right) = \frac{e}{2} - \frac{3}{2} \approx \boxed{-0.140859086}$

Let $\displaystyle{G(u,v,x)=\int_{v}^{u}{e^{{t^2}+tx}}dt \implies F(x)=G(cos(x),sin(x),x)}$

Then $\displaystyle{F'(x)=\frac{\partial G}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial G}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial G}{\partial x}\frac{\partial x}{\partial x}}$

$\displaystyle{=e^{{u^2}+ux}(-sin(x))-e^{{v^2}+vx}(cos(x))+\int_{v}^{u}{te^{{t^2}+tx}}dt}$

$u(x)=cos(x)\Rightarrow u(0)=1$

$v(x)=sin(x)\Rightarrow v(0)=0$

So $\displaystyle{ F'(0)=e^{1^2+1(0)}(-0)-e^{0^2+0(0)}(1)+\int_0^1{te^{t^{2}+t(0)}dt }}$

$\displaystyle{=-1+\int_0^1{te^{t^2}dt} = \frac{e-3}{2} }$