Here are a few of my favourites
Integration by cancellation
Assume you are to integrate some function that can be written as
the product of two functions, $f = g \cdot h$. The idea now is to
use integration by parts on $g$ such that the integral over $h$ disappears.
Example: Let $f(x) = (1 + 2x^2) e^{x^2}$, most techniques will not work here.
Give it a go with integration by parts or any substitution you like. The "trick" however
is to split the integral
\begin{align*}
J
= \int (1 + 2x^2) e^{x^2} \mathrm{d}x
= \int 2x^2e^{x^2}\mathrm{d}x + \int e^{x^2} \mathrm{d}x \,,
\end{align*}
and use integration by parts on the last integral with $u = e^{x^2}$ and $v=x$.
So
\begin{align*}
J
= \int 2x^2e^{x^2} \mathrm{d}x + \left[ x e^{x^2}
- \int x \cdot 2x e^{x^2} \mathrm{d}x \right]
= x e^{x^2} + \mathcal{C}
\end{align*}
This is nothing else than using the product rule backwards, however I often
find it easier to look at this way.
$$ \int \log( \log x ) - \frac{\mathrm{d}x}{\log x} $$
Integration over symmetric functions
(Roger Nelsen) Let $f$ be a bounded function on $[a,b]$ then
\begin{align*}
\int_a^b f(x) = (b-a) f\left( \frac{a+b}{2} \right)
= \frac{b-a}{2}\bigl[ f(a) + f(b)\bigr]
\end{align*}
given that $f(x)+f(a+b-x)$ is constant for all $x\in[a,b]$
$$
\int_0^{\pi/2} \frac{\mathrm{d}x}{1 + \tan(x)^{\sqrt{2}}}
$$
Integration over periodic functions
Let $f$ be a function such that $f(x) = f(x+T)$ for all $x$, with $T \in \mathbb{R}$
then
\begin{align}
\int_{a}^{a+T} f(x)\,\mathrm{d}x & = \phantom{k}\int_{b}^{b + T} f(x)\,\mathrm{d}x\\
\int_{0}^{kT\phantom{a}} f(x)\,\mathrm{d}x & = k \int_0^T f(x)\,\mathrm{d}x\\
\int_{a + mT}^{b + nT} f(x)\,\mathrm{d}x & = \int_a^bf(x)\,\mathrm{d}x+(n-m)\int_0^{T} f(x)\,\mathrm{d}x\,
\end{align}
where $a,b,k,n,m$ are real numbers
$$
\int_{23\pi}^{71\pi/2} \frac{\mathrm{d}x}{1 + 2^{\sin x}}
$$
Functional equation
Let $R(x)$ be some rational function satisfying
\begin{align*}
R\left(\frac{1}{x}\right) \frac{1}{x^2} = R(x)\,,
\end{align*}
for all $x$. Then
\begin{alignat}{2}
& \int_0^\infty R(x) \,\mathrm{d}x && = \;2 \int_0^1 R(x) \\
& \int_0^\infty R(x) \log x \,\mathrm{d}x && = \;0 \\
& \int_0^\infty \frac{R(x)}{x^b + 1} \,\mathrm{d}x && = \frac{1}{2} \int_0^\infty R(x) \,\mathrm{d}x\\
& \int_0^\infty R(x) \arctan x \,\mathrm{d}x && = \frac{\pi}{4} \int_0^\infty R(x) \,\mathrm{d}x
\end{alignat}
$$
\int_0^{\pi/2} \frac{\log ax}{b^2+x^2} \mathrm{d}x
$$
More of these identities can be found for an example here with proofs.
