Biot Savart - Vortex line segment parallel to a plane I am trying to understand what is shown in this book (Applied hydro and aeromechanics), page 187
https://books.google.com.au/books?id=Ds-bd0zAwIYC
The velocity induced at a given point on a parallel plane by a vortex segment of length $b$, and strength $\Gamma$ is shown to be,
$v_1 = \frac{\Gamma b \, sin \, \alpha}{4 \pi R^2}$
Can anyone please show the steps on arriving this formula by applying biot-savart low ?
Thanks much in advance
ABCD
 A: For reference, here is the figure:

We are interested in the influence of the wing-bound vortex of length $b$ and circulation $\Gamma$ on point $A$, particularly the velocity vector $V_1$.
The Biot-Savart law in this case can be written as
\begin{equation}
V_1=\frac{\Gamma}{4\pi}\int\frac{\mathrm{d}\vec{b}\times\vec{R}}{|\vec{R}|^3}, \tag{1}
\end{equation}
where $\vec{R}$ is the vector with magnitude $R$ in the figure and $\mathrm{d}\vec{b}$ is an infinitesimal section of the vector representation of the bound vortex. Please see Karamcheti for a full derivation; it's rather too involved for us here.
Using $b_1$ and $b_2$ as the ends of the bound vortex, we have
\begin{equation}
V_1=\frac{\Gamma}{4\pi}\int_{b_1}^{b_2}\frac{\mathrm{d}\vec{b}\times\vec{R}}{|\vec{R}|^3}.  \tag{2}
\end{equation}
We can see from the figure the direction of $V_1$ and a simple application of a right-hand rule on the bound vortex confirms this orientation. We can thus focus on the magnitude of $V_1$, which is $v_1$.
To simplify, we can use the definition of the cross product:
\begin{equation}
\mathrm{d}\vec{b}\times\vec{R}=\mathrm{d}bR\sin{\alpha}, \tag{3}
\end{equation}
which goes back into eq. (2), leaving us with
\begin{equation}
v_1=\frac{\Gamma}{4\pi}\int_{b_1}^{b_2}\frac{R\sin{\alpha}}{R^3}\mathrm{d}b =\frac{\Gamma}{4\pi}\int_{b_1}^{b_2}\frac{\sin{\alpha}}{R^2}\mathrm{d}b. \tag{4}
\end{equation}
When we integrate over the length of $b$, $\mathrm{d}b$ just becomes $b$. We thus wind up with
\begin{equation}
v_1=\frac{\Gamma b \sin{\alpha}}{4\pi R^2}.
\end{equation}
