It sounds as if the professor is talking about contour integration. If complex analysis is a prerequisite for this course, then that is most likely what it is.
An example of an integral with a real singularity and a removable singularity is the principal value integral $ \mathrm{PV}\int_{-\infty}^\infty\frac{e^{ix}}{x}\mathrm{d}x =\lim_{R\to\infty}\left(\int_{-R}^{-1/R}\frac{e^{ix}}{x}\mathrm{d}x+\int_{1/R}^{R}\frac{e^{ix}}{x}\mathrm{d}x\right)\tag{1} $ Since $e^{ix}=\cos(x)+i\sin(x)$, $(1)$ covers a principal value integral with a real singularity and an integral with a removable singularity: $ \mathrm{PV}\int_{-\infty}^\infty\frac{\cos(x)}{x}\mathrm{d}x\qquad\text{and}\qquad\int_{-\infty}^\infty\frac{\sin(x)}{x}\mathrm{d}x\tag{2} $ The integral on the left hand side of $(1)$ can be handled with the following contour:
$\hspace{3.2cm}$
The integral in $(1)$ equals the integral of $\dfrac{e^{iz}}{z}$ over the two red pieces of the contour above as the large curve (radius $R$) gets larger and the small curve (radius $1/R$) gets smaller. To use contour integration, we close the contour with the two circular curves.
Since there is no singularity of inside the contour, the integral over the entire contour is $0$.
The integral over the large curve where $y<\sqrt{R}$ tends to $0$ since the absolute value of the integrand is $\le1/R$ over two pieces of the contour whose length is essentially $\sqrt{R}$. The integral over the large curve where $y\ge\sqrt{R}$ also tends to $0$ since the absolute value of the integral is less than $e^{-\sqrt{R}}/R$ over a contour whose length is less than $\pi R$. Thus, the integral over the large curve tends to $0$.
Near the origin, we have $ \frac{e^{iz}}{z}=\frac1z+i-\frac{z}{2}-i\frac{z^2}{6}+\dots\tag{3} $ Aside from $\frac1z$, the series in $(3)$ is bounded and so its integral over the small curve tends to $0$ since the length of the curve is $\pi/R$. The integral of $\frac1z$ over a counter-clockwise circular arc centered at the origin is $i$ times the angle of the arc. Thus, the integral over the small curve is $-\pi i$ (clockwise circular arc with angle $\pi$).
Since the integral over the entire contour is $0$ and the integral over the two curves tends to a total of $-\pi i$, the integral over the two red pieces must tend to $\pi i$. Taking real and imaginary parts yields $ \mathrm{PV}\int_{-\infty}^\infty\frac{\cos(x)}{x}\mathrm{d}x=0\qquad\text{and}\qquad\int_{-\infty}^\infty\frac{\sin(x)}{x}\mathrm{d}x=\pi\tag{4} $ Alternatively, we can also use the following contour:
$\hspace{3.2cm}$
Everything is as above except that the contour contains the singularity at $0$ and the small circular curve is counter-clockwise.
The integral over the entire contour is $2\pi i$ since the residue of the singularity is $1$.
The integral over the large curve tends to $0$ as above.
The integral over the small curve is $\pi i$ (counter-clockwise circular arc with angle $\pi$).
Therefore, the integral over the two red pieces must again tend to $\pi i$.
Thus, the integral can be handled circling the singularity either way.
Perhaps something like this is what is going on in class.