Integral From Electricity and Magnetism In attempting to calculate the magnetic field due to a current loop at an arbitrary point, I ran into the following integrals
\begin{align}
&\int_0^{2\pi} \frac{z\cos t \ dt}{\left[ \cos^2 t+ (y - \sin t)^2 + z^2\right]^{\frac{3}{2}}}
\\&\int_0^{2\pi} \frac{z\sin t \ dt}{\left[ \cos^2 t+ (y - \sin t)^2 + z^2\right]^{\frac{3}{2}}}
\\&\int_0^{2\pi} \frac{1 - y\sin t \ dt}{\left[ \cos^2 t+ (y - \sin t)^2 + z^2\right]^{\frac{3}{2}}}
\end{align}
I've set $x = 0$ without loss of generality (there is rotational symmetry about the $z$ axis). The integrals are the $x$, $y$, and $z$ components of the magnetic field respectively with all physical constants set to $1$. Are these well-known integrals? They remind me of elliptic integrals, so I don't really expect a closed-form solution. It's often claimed that a current loop is a magnetic dipole, so, in principle, these integrals should look like a dipole field for large $y$ and $z$. Is there an easy way to see this?
 A: I will work with integrals without normalization. Write $a$ for the diameter of the current loop. Then the first integral is simply
\begin{align*}
\int_{0}^{2\pi} \frac{a z \cos t}{(a^2 \cos^2 t + (y-a \sin t)^2 + z^2)^{3/2}} \, \mathrm{d}t
&= 0.
\end{align*}
The other two integrals involve elliptic functions, but its asymptotic expansion far away from the origin can be computed easily as well. To this end, write $r = \sqrt{y^2 + z^2}$ for the distance from the origin. Then from the Taylor expansion, we have
\begin{align*}
\frac{1}{(a^2 \cos^2 t + (y-a \sin t)^2 + z^2)^{3/2}}
&= \frac{1}{r^{3}} + \frac{3a y \sin t}{r^5} + \mathcal{O}\left(\frac{a^2}{r^5}\right)
\end{align*}
for $a \ll r$. Using this, we get
\begin{align*}
&\int_{0}^{2\pi} \frac{a z \sin t}{(a^2 \cos^2 t + (y-a \sin t)^2 + z^2)^{3/2}} \, \mathrm{d}t \\
&\quad= \int_{0}^{2\pi} \left( \frac{1}{r^{3}} + \frac{3a y \sin t}{r^5} + \mathcal{O}\left(\frac{a^2}{r^5}\right) \right) (a z \sin t) \, \mathrm{d}t \\
&\quad = \frac{3a^2 \pi y z}{r^5} + \mathcal{O}\left(\frac{a^3}{r^4}\right)
\end{align*}
and
\begin{align*}
&\int_{0}^{2\pi} \frac{a(a - y \sin t)}{(a^2 \cos^2 t + (y-a \sin t)^2 + z^2)^{3/2}} \, \mathrm{d}t \\
&\quad= \int_{0}^{2\pi} \left( \frac{1}{r^{3}} + \frac{3a y \sin t}{r^5} + \mathcal{O}\left(\frac{a^2}{r^5}\right) \right) a (a - y \sin t) \, \mathrm{d}t \\
&\quad= \frac{3a^2\pi z^2}{r^5} - \frac{\pi a^2}{r^3} + \mathcal{O}\left(\frac{a^3}{r^4}\right).
\end{align*}
Now by writing $\mathbf{r} = y \hat{\mathbf{y}} + z\hat{\mathbf{z}}$ for the the displacement and $\mathbf{m}=I \pi a^2 \hat{\mathbf{z}}$ for the magnetic moment (where $I$ is the current on the loop), the magnetic field $\mathbf{B}$ takes the following asymptotic form
\begin{align*}
\mathbf{B}
&= \frac{\mu_0 I}{4\pi} \left[ \frac{3a^2 \pi y z}{r^5} \hat{\mathbf{y}} + \left( \frac{3a^2\pi z^2}{r^5} - \frac{\pi a^2}{r^3} \right) \hat{\mathbf{z}} + \mathcal{O}\left(\frac{a^3}{r^4}\right) \right] \\
&= \frac{\mu_0}{4\pi} \left[ \frac{3\mathbf{r} (\mathbf{m} \cdot \mathbf{r})}{r^5} - \frac{\mathbf{m}}{r^3} \right] + \mathcal{O}\left(\frac{a^3}{r^4}\right)
\end{align*}
The leading term is exactly the formula described in Wikipedia article.
