Answer
$$
x=\cos t+\ln \left(\tan \frac{1}{2} t\right), \quad y=\sin t, \quad\pi / 4 \leq t \leq 3 \pi / 4
$$
the exact length of the curve is
$$
\begin{aligned} L &=\int_{\pi / 4}^{3 \pi / 4} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=\ln 2 \end{aligned}
$$
Work Step by Step
$$
x=\cos t+\ln \left(\tan \frac{1}{2} t\right), \quad y=\sin t, \quad\pi / 4 \leq t \leq 3 \pi / 4
$$
then
$$
\begin{aligned}
\frac{d x}{d t}&=-\sin t+\frac{\frac{1}{2} \sec ^{2}(t / 2)}{\tan (t / 2)}\\
&=-\sin t+\frac{1}{2 \sin (t / 2) \cos (t / 2)} \\
&=-\sin t+\frac{1}{\sin t},
\end{aligned}
$$
and
$$
\frac{d y}{d t}=\cos t
$$
so
$$
\begin{aligned}
\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}&=\sin ^{2} t-2+\frac{1}{\sin ^{2} t}+\cos ^{2} t \\
&=1-2+\csc ^{2} t \\
&=\cot ^{2} t
\end{aligned}
$$
Thus the exact length of the curve is
$$
\begin{aligned} L &=\int_{\pi / 4}^{3 \pi / 4} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=\int_{\pi / 4}^{3 \pi / 4}|\cot t| d t=2 \int_{\pi / 4}^{\pi / 2} \cot t d t \\ &=2[\ln |\sin t|]_{\pi / 4}^{\pi / 2}=2\left(\ln 1-\ln \frac{1}{\sqrt{2}}\right) \\ &=2(0+\ln \sqrt{2})=2\left(\frac{1}{2} \ln 2\right) \\
&=\ln 2 \end{aligned}
$$
