We first compute \(\|\vect\gamma(t)\|=\sqrt{\bigl(\gamma_1'(t)\bigr)^2+\bigl(\gamma_2'(t)\bigr)^2}\text{.}\) As \(\gamma_1'(t)=1\) and \(\gamma_2'(t)=\sinh t\) we get
\begin{equation*}
\|\vect\gamma'(t)\|
=\sqrt{1+\sinh^2 t}=\sqrt{\cosh^2 t}=\cosh t\text{.}
\end{equation*}
Hence,
\begin{align*}
\int_C(x^2-y)\,ds
\amp=\int_0^1((\gamma_1(t))^2-\gamma_2(t))
\sqrt{(\gamma_1'(t))^2+(\gamma_2'(t))^2)}\,dt\\
\amp=\int_0^1(t^2-\cosh t)\cosh t\,dt\\
\amp=\int_0^1t^2\cosh t\,dt-\int_0^1\cosh^2 t\,dt
\end{align*}
We now compute the last two integrals separately. For the first we integrate by parts twice:
\begin{align*}
\int_0^1t^2\cosh t\,dt
=\bigl[t^2\sinh t\bigl]_0^1-2\int_0^1t\sinh t\,dt
\amp=\sinh 1-2\Bigl(\bigl[t\cosh t\bigl]_0^1-\int_0^1\cosh t\,dt\Bigr)\\
\amp=\sinh 1-2\cosh 1+2\bigl[\sinh t\bigl]_0^1\\
\amp=\sinh 1-2\cosh 1+2\sinh 1=3\sinh 1-2\cosh 1\text{.}
\end{align*}
Using a well known identity and integration of parts we have
\begin{align*}
\int\cosh^2 t\,dt
\amp=\sinh t\cosh t-\int\sinh^2 t\,dt\\
\amp=\sinh t\cosh t-\int(\cosh^2t-1)\,dt\\
=\sinh t\cosh t+t-\int\cosh^2 t\,dt\text{.}
\end{align*}
Hence,
\begin{equation*}
\int_0^1\cosh^2 t\,dt
=\frac{1}{2}\Bigl[t+\sinh t\cosh t\Bigr]_0^1
=\frac{1+\sinh 1\cosh 1}{2},
\end{equation*}
and therefore
\begin{equation*}
\int_C(x^2-y)\,ds
=3\sinh 1-2\cosh 1-\frac{1+\sinh 1\cosh 1}{2}\text{.}
\end{equation*}
This is the required result.
