Detail from Integral Evaluation in Szabo (Appendix) This is a step from Appendix A in Szabo and Ostlund's Quantum Chemistry. In short, the question is how to get from \ref{A.20} to \ref{A.21}. We can factor out some constants $K$ in the expressions to get:

\begin{align}
  (A| -Z_C/r_{1C} |B) 
  &amp;=
    K \left(\frac{1}{2\pi^2}\right)
    \int \mathrm{d}\mathbf{k} \; e^{-k^2/4p} k^{-2}
    \exp\big(-i\,\mathbf{k}\cdot(\mathbf{R}_P - \mathbf{R}_C)\big)
    \tag{A.20}\label{A.20}\\
  &amp;=
    K \left(\frac{2}{\pi|\mathbf{R}_P-\mathbf{R}_C|}\right)
    \int\limits_0^\infty \mathrm{d}k \; e^{-k^2/4p} k^{-1} 
    \sin(k|\mathbf{R}_P-\mathbf{R}_C|)
    \tag{A.21}\label{A.21}
\end{align}

Part of the note between these steps is that (paraphrasing) we can let $\mathbf{R}_P-\mathbf{R}_C$ lie on the $z$-axis and use that $k\cdot (\mathbf{R}_P-\mathbf{R}_C)= k|\mathbf{R}_P-\mathbf{R}_C|\cos\theta$, "then we can easily perform the angular part of the integration over $\boldsymbol{k}$ to obtain [\ref{A.21}]."  

What I have done: It seems that Euler's formula must be applied to the second exponent in \ref{A.20} but using the hint we'd have something like $$\cos(k|\mathbf{R}_P-\mathbf{R}_C|\cos\theta) + i\sin \dots \text{etc}. \tag{???}$$
 I guess that the sine in \ref{A.21} follows integration of the (real) cosine in Euler's expansion. In line with this, \ref{A.21} is multiplied by a factor of $2\pi$.

I will keep at this but I think if it puzzles me, then others may have stumbled over it so I am posting.
Hopefully removal of the common factors ($K$) makes this easier to read but the problem is on pages 413-414 of Szabo and Ostlunds book.
As commented, do note that the integral (A.20) is taken over all $\mathbf{k}$-space, whereas the integral (A.21) is one-dimensional. The explicit calculation of this integral follows as 

$$\begin{align} \int\text{d}\mathbf{k}\,\Psi(k)e^{-i\mathbf{k\cdot r}} &amp;=\int_0^\infty\text{d}k\,k^2\int_0^\pi\text{d}\theta\int_0^{2\pi}\text{d}\varphi\,\sin\theta\Psi(k)e^{-ikr\cos\theta}\\
&amp;= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\int_0^\pi\text{d}\theta\,\sin\theta e^{-ikr\cos\theta}\\
&amp;= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\left[\frac{e^{ikru}}{ikr}\right]_{u=-1}^1\\
&amp;= \frac{4\pi}{r}\int_0^\infty\text{d}k\,k\Psi(k)\sin(kr).\end{align}$$

In the first equality we convert this integral from Cartesian to spherical coordinates, in the second equality we evaluate the integral over $\varphi$, in the third equality we substitute $u = -\cos\theta$, and in the fourth equality we make use of Euler's formula to get the exponential representation of $\sin(kr)$.