For convenience I use $x$ as the variable of integration. Let
$$
F(t)=I(u(t),v(t),t)=\int_{u(t)}^{v(t)}f(x,t)dx
$$
You have variable limits and need to differentiate under the integral sign. By the Leibniz integral rule, the derivative is
$$\begin{eqnarray*}
F^{\prime }(t) &=&\frac{\partial I}{\partial t}\frac{dt}{dt}+\frac{\partial I
}{\partial v}\frac{dv}{dt}+\frac{\partial I}{\partial u}\frac{du}{dt} \\
&=&\int_{u(t)}^{v(t)}\frac{\partial f(x,t)}{\partial t}dx+f(v(t),t)v^{\prime
}(t)-f(u(t),t)u^{\prime }(t).
\end{eqnarray*}$$
Hence
$$
F^{\prime }(0)=\int_{u(0)}^{v(0)}\left. \frac{\partial f(x,t)}{\partial t}
\right\vert _{t=0}dx+f(v(0),0)v^{\prime }(0)-f(u(0),0)u^{\prime }(0).
$$
In the present case
$$
\begin{eqnarray*}
u(t) &=&\sin t,\qquad u(0)=0 \\
u^{\prime }(t) &=&\cos t,\qquad u^{\prime }(0)=1 \\
v(t) &=&\cos t,\qquad v(0)=1 \\
v^{\prime }(t) &=&-\sin t,\qquad v^{\prime }(0)=0.
\end{eqnarray*}$$
and
$$
\begin{eqnarray*}
f(x,t) =e^{x^{2}+xt}, \qquad f(1,0) =e, \qquad f(0,0) =1.
\end{eqnarray*}$$
So
$$
F^{\prime }(0)=\int_{0}^{1}\left. \frac{\partial f(x,t)}{\partial t}
\right\vert _{t=0}dx+0-1
$$
Since
$$
\left. \frac{\partial f(x,t)}{\partial t}\right\vert _{t=0}=\left. \frac{
\partial }{\partial t}e^{x^{2}+xt}\right\vert _{t=0}=xe^{x\left( x+t\right)
}\vert _{t=0}=xe^{x^{2}}
$$
and
$$
\int_{0}^{1}xe^{x^{2}}dx=\frac{1}{2}e-\frac{1}{2}
$$
we finally get
$$
F^{\prime }(0)=\frac{1}{2}e-\frac{1}{2}-1=\frac{1}{2}e-\frac{3}{2}.
$$
You find another example here.
