Answer
$$\frac{3 \pi^{2}}{8}$$
Work Step by Step
Since
$$\mathbf{r}=(\cos t+t \sin t) \mathbf{i}+(\sin t-t \cos t) \mathbf{j}$$Then \begin{align*}
\mathbf{v}&=(-\sin t+\sin t+t \cos t) \mathbf{i}+(\cos t-\cos t+t \sin t) \mathbf{j}\\
&=(t \cos t) \mathbf{i}+(t \sin t) \mathbf{j} \\
|\mathbf{v}|&=\sqrt{(t \cos t)^{2}+(t \cos t)^{2}}\\
&=\sqrt{t^{2}}=t,\end{align*}
Hence
\begin{align*}
\Rightarrow s(t)&=\int_{0}^{t} \tau d \tau\\
&=\frac{t^{2}}{2}\\
\text{Length }&=s(\pi)-s\left(\frac{\pi}{2}\right)\\
&=\frac{\pi^{2}}{2}-\frac{\left(\frac{\pi}{2}\right)^{2}}{2}\\
&=\frac{3 \pi^{2}}{8}
\end{align*}
