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.