While back, I was calculating a physics problem involving an integral of the form
$$\int\frac{dx}{\sqrt{x^2+a}+b}$$
Naturally, one would begin with the trig substitution $x=a\tan\theta$. So, the integral can be written as
\begin{align*} \int\frac{dx}{\sqrt{x^2+a}+b}&=\int\frac{a\sec^2\theta d\theta}{a\sec\theta+b}\\ &=\frac{1}{a}\int\frac{a^2\sec^2\theta-b^2+b^2}{a\sec\theta+b}d\theta\\ &=\int\sec\theta d\theta-\frac{b}{a}\theta+\frac{b^2}{a}\int\frac{d\theta}{a\sec\theta+b}\\ &=\ln|\sec\theta+\tan\theta|-\frac{b}{\sqrt{b^2-a^2}}\ln\frac{\sqrt{b^2-a^2}\tan\left(\frac{\theta}{2}\right)+a+b}{\sqrt{b^2-a^2}\tan\left(\frac{\theta}{2}\right)-(a+b)}\\ &=\ln\left|\frac{\sqrt{x^2+a^2}+x}{a}\right|-\frac{b}{\sqrt{b^2-a^2}}\ln\left[\frac{\frac{\sqrt{b^2-a^2}x}{a+\sqrt{x^2+a^2}}+a+b}{\frac{\sqrt{b^2-a^2}x}{a+\sqrt{x^2+a^2}}-(a+b)}\right] \end{align*}
Here,
\begin{align*} \int\frac{d\theta}{a\sec\theta+b}&=\int\frac{\cos\theta d\theta}{a+b\cos\theta}\\ &=\frac{1}{b}\int d\theta-\frac{a}{b}\int\frac{d\theta}{a+b\cos\theta}\\ &=\frac{1}{b}\theta-\frac{a}{b}\frac{1}{\sqrt{b^2-a^2}}\ln\frac{\sqrt{b^2-a^2}\tan\left(\frac{\theta}{2}\right)+a+b}{\sqrt{b^2-a^2}\tan\left(\frac{\theta}{2}\right)-(a+b)} \end{align*}
For the evaluation of $\int\frac{d\theta}{a+b\cos\theta}$, I used the formula from here.

The Gaussian integral is not only important in mathematics but is also extremely important in studying physics, for example, quantum mechanics, quantum field theory, and statistical mechanics. Let us begin with the most simple Gaussian integral
$$G=\int_{-\infty}^\infty e^{-x^2}dx$$
This is not an easy integral to calculate but it can be done easily if we extend it to an integral in two-dimensions as
\begin{align*} G^2&=\int_{-\infty}^\infty e^{-x^2}dx\int_{-\infty}^\infty e^{-y^2}dy\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}dxdy \end{align*}
Now, in term of the polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$ and the area element is $dA=rdrd\theta$, so $G^2$ can be written as
\begin{align*} G^2&=\int_0^{2\pi}\int_0^\infty re^{-r^2}drd\theta\\ &=\pi \end{align*}
The integral $\int_0^\infty re^{-r^2}dr$ can be easily evaluated using the substitution $u=-r^2$ and its value is $\frac{1}{2}$. Now, we obtain
$$G=\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}$$
From this you can evaluate more complicated Gaussian integrals using a simple substitution and/or some algebra. For example, $\int_{-\infty}^\infty e^{-ax^2}dx$ for a positive real number $a$ can be evaluated easily by a simple substitution $u=\sqrt{a}x$:
\begin{equation}
\begin{aligned}
\int_{-\infty}^\infty e^{-ax^2}dx&=\frac{1}{\sqrt{a}}\int_{-\infty}^\infty e^{-u^2}du\\
&=\sqrt{\frac{\pi}{a}}
\end{aligned}\label{eq:gaussint}
\end{equation}
If $a=\frac{1}{2}$, then $\int_{-\infty}^\infty e^{-\frac{1}{2}x^2}dx=\sqrt{2\pi}$. Normally one will have to use the integration by parts to calculate $\int_{-\infty}^\infty x^2e^{-x^2}dx$ or $\int_{-\infty}^\infty x^4e^{-x^2}dx$. There is a neat trick of evaluating such integrals by differentiating \eqref{eq:gaussint} with respect to $a$:
$$\frac{d}{da}\int_{-\infty}^\infty e^{-ax^2}dx=-\int_{-\infty}^\infty x^2e^{-ax^2}dx$$
and
$$\frac{d}{da}\sqrt{\frac{\pi}{a}}=-\frac{1}{2a}\sqrt{\frac{\pi}{a}}$$
Thus, we have
\begin{equation}
\label{eq:gaussint2}
\int_{-\infty}^\infty x^2e^{-ax^2}dx=\frac{1}{2a}\sqrt{\frac{\pi}{a}}
\end{equation}
For $a=1$, we obtain
$$\int_{-\infty}^\infty x^2e^{-x^2}dx=\frac{\sqrt{\pi}}{2}$$
Differentiating \eqref{eq:gaussint2} with respect to $a$ leads to
\begin{equation}
\label{eq:gaussint3}
\int_{-\infty}^\infty x^4e^{-ax^2}dx=\frac{3}{4a^2}\sqrt{\frac{\pi}{a}}
\end{equation}
For $a=1$, we obtain
\begin{equation}
\label{eq:gaussint4}
\int_{-\infty}^\infty x^4e^{-x^2}dx=\frac{3}{4}\sqrt{\pi}
\end{equation}
I learned this trick from the book Quantum Field Theory in a Nutshell by Anthony Zee. Another important variation is
\begin{equation}
\label{eq:gaussint5}
\int_{-\infty}^\infty e^{-ax^2+bx}dx
\end{equation}
This integral can be evaluated from \eqref{eq:gaussint} with a little bit of algebra. First, by completing the square, we write
\begin{align*} -ax^2+bx&=-a\left(x^2-\frac{b}{a}x\right)\\ &=-a\left(x^2-\frac{b}{a}x+\left(-\frac{b}{2a}\right)^2-\left(-\frac{b}{2a}\right)^2\right)\\ &=-a\left(x-\frac{b}{2a}\right)^2+\frac{b^2}{4a} \end{align*}
Now, \eqref{eq:gaussint5} is evaluated as
\begin{align*} \int_{-\infty}^\infty e^{-ax^2+bx}dx&=\int_{-\infty}^\infty e^{-a\left(x-\frac{b}{2a}\right)^2+\frac{b^2}{4a}}dx\\ &=e^{\frac{b^2}{4a}}\int_{-\infty}^\infty e^{-a\left(x-\frac{b}{2a}\right)^2}dx\\ &=e^{\frac{b^2}{4a}}\int_{-\infty}^\infty e^{-au^2}du\ \left[u=x-\frac{b}{2a}\right]\\ &=e^{\frac{b^2}{4a}}\sqrt{\frac{\pi}{a}} \end{align*}

This time let us try to calculate
\begin{equation}
\label{eq:gaussint6}
\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty (x^2+y^2+z^2)e^{-(x^2+y^2+z^2)}dxdydz
\end{equation}
You stumble upon this type of integral, for example, in a derivation of the Maxwell-Boltzmann statistics in the form of
$$\int d^3p\frac{p^2}{2m}\exp\left(-\beta\frac{p^2}{2m}\right)$$
The integral \eqref{eq:gaussint6} can be easily evaluated using the spherical coordinates:
\begin{align*} x&=r\sin\theta\cos\phi\\ y&=r\sin\theta\sin\phi\\ z&=r\cos\theta\ \end{align*}
where
$$0\leq r<\infty,\ 0\leq\theta\leq\pi,\ 0\leq\theta\leq 2\pi$$
The volume element in the spherical coordinates is $dV=r^2dr\sin\theta d\theta d\phi$, so
\begin{align*} \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty (x^2+y^2+z^2)e^{-(x^2+y^2+z^2)}dxdydz&=\int_0^{2\pi}\int_0^\pi\int_0^\infty r^4e^{-r^2}dr\sin\theta d\theta d\phi\\&=\frac{3}{2}\pi\sqrt{\pi} \end{align*}
We obtain
$$\int_0^\infty r^4e^{-r^2}dr=\frac{3}{8}\sqrt{\pi}$$
using \eqref{eq:gaussint4}.

One of my Ph.D. students, Victor Zankoni is reading the paper Relativistically corrected Schrödinger equation with Coulomb interaction by J. L. Friar and E. L. Tomusiak, Physical Review C, Volume 29, Number 4, 1537. Following their calculations, he stumbled upon an integral of the form
$$\int\frac{dx}{a+b\cos x}$$
where $b^2>a^2$. Thinking it may not be easy to calculate, I suggested him to look up the well-known book Tables of Integrals, Series, and Products by Gradshteyn and Ryzhik. He said he couldn’t find a relevant formula (actually there is and he somehow missed it) and instead he decided to calculate it. He used a clever substitution and it worked beautifully, so here I am introducing his calculation. First note the identity
$$\cos x=\frac{1-\tan^2\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)}$$
Here, I assume $a>0$ and $b>0$ but similar calculations can be carried out for other cases as well. With the above substitution, we have
\begin{align*} \int\frac{dx}{a+b\cos x}&=\int\frac{\sec^2\left(\frac{x}{2}\right)}{b+a-(b-a)\tan^2\left(\frac{x}{2}\right)}dx\\ &=\frac{2}{b-a}\int\frac{du}{\left(\sqrt{\frac{b+a}{b-a}}\right)^2-u^2}\\ &=\frac{1}{\sqrt{b^2-a^2}}\left[\int\frac{du}{\sqrt{\frac{b+a}{b-a}}+u}+\int\frac{du}{\sqrt{\frac{b+a}{b-a}}-u}\right]\\ &=\frac{1}{\sqrt{b^2-a^2}}\ln\frac{u+\sqrt{\frac{b+a}{b-a}}}{u-\sqrt{\frac{b+a}{b-a}}}\\ &=\frac{1}{\sqrt{b^2-a^2}}\ln\frac{\sqrt{b^2-a^2}u+a+b}{\sqrt{b^2-a^2}u-(a+b)}\\ &=\frac{1}{\sqrt{b^2-a^2}}\ln\frac{\sqrt{b^2-a^2}\tan\left(\frac{x}{2}\right)+a+b}{\sqrt{b^2-a^2}\tan\left(\frac{x}{2}\right)-(a+b)} \end{align*}
The last expression coincides with 2.553 #3 of Gradshteyn and Ryzhik, 8th Edition on page 172. Gradshteyn and Ryzhik 7th Edition does not contain this expression but one can find a supposedly equivalent expression in 2.553 #3 of the 7th Edition but unfortunately, it is stated incorrectly. It has been corrected in the 8th Edition and is listed as 2.553 #4:
$$\int\frac{dx}{a+b\cos x}=\frac{2}{\sqrt{b^2-a^2}}\ln\left|\frac{(b-a)\tan\left(\frac{x}{2}\right)+\sqrt{b^2-a^2}}{(b-a)\tan\left(\frac{x}{2}\right)-\sqrt{b^2-a^2}}\right|\ [b^2>a^2]$$
There is another incorrect formula that caught my eyes:
\begin{align*} \int\frac{dx}{a+b\cos x}&=\frac{2}{\sqrt{b^2-a^2}}\mathrm{arctanh}\left(\frac{(a-b)\tan\left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)\ \left[b^2>a^2,\ \left|(b-a)\tan\left(\frac{x}{2}\right)\right|<\sqrt{b^2-a^2}\right]\\ &=\frac{2}{\sqrt{b^2-a^2}}\mathrm{arccoth}\left(\frac{(a-b)\tan\left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)\ \left[b^2>a^2,\ \left|(b-a)\tan\left(\frac{x}{2}\right)\right|>\sqrt{b^2-a^2}\right] \end{align*}
in both 7th Edition and the 8th Edition. It should be corrected to
\begin{align*} \int\frac{dx}{a+b\cos x}&=\frac{2}{\sqrt{b^2-a^2}}\mathrm{arctanh}\left(\frac{(b-a)\tan\left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)\ \left[b^2>a^2,\ \left|(b-a)\tan\left(\frac{x}{2}\right)\right|<\sqrt{b^2-a^2}\right]\\ &=\frac{2}{\sqrt{b^2-a^2}}\mathrm{arccoth}\left(\frac{(b-a)\tan\left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)\ \left[b^2>a^2,\ \left|(b-a)\tan\left(\frac{x}{2}\right)\right|>\sqrt{b^2-a^2}\right] \end{align*}
I don’t use Gradshteyn and Ryzhik’s book much but I have heard that it contains numerous typos and mistakes. It appears to be true. So, please use it with caution when you do.