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I'm learning to integrate and I'd like to hear what are you favorite integration tricks?

I can't contribute much to this thread, but I like the fact that:

$$\int_{-a}^{a}{f(x)}dx=0 \space\text{if}\space f(x) \space\text{is odd}$$

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Perhaps this need to be CW. – Joe Jun 23 '12 at 20:56
Users are no longer allowed to make their own questions CW (this is to discourage CW questions, which are not really the point of StackExchange). But you can flag for moderator attention and a moderator can do it. – Qiaochu Yuan Jun 23 '12 at 21:13
There's a list of good and somewhat exotic integration tricks in this thread on "lesser-known" integration tricks. Some very common integrals are collected in the abstract duplicate questions thread on meta under the heading "Calculus". – t.b. Jun 23 '12 at 21:21
@ncmathsadist: Community Wiki [mode]; also Choice Weirdos, Celestial World, Covert Wallaby, Crystal Wardrobe, Creeping Weasel, Cool Whatchamaccallit, Christian Wars, etc. – Asaf Karagila Jun 23 '12 at 22:14
One of my favourite tricks for finding an antiderivative of $f(x)$ is making a reasonable guess $F(x)$, differentiating, and seeing what adjustment to $F(x)$ will make the derivative be $f(x)$. – André Nicolas Jun 24 '12 at 4:02
up vote 2 down vote accepted

Integration by parts.

There is even a separate thread on stack-exchange on integration by parts.

Striking applications of integration by parts

The discrete analog of integration by parts i.e. the summation by parts is also an important tool especially in analytical number theory when we want to find asymptotic. For instance, my recent post here, uses this to get an estimate of $\displaystyle \sum_{n=N+1}^{\infty} \dfrac1{n^s}$.

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  1. Residues. They're fantastic. One small part of this is the evaluation of a rational function of $\sin(x), \cos(x)$. By making a simple substitution, you can reduce such integrals to 'ordinary' ones, and then use residues.
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Could you add an example? – bodacydo Jul 21 '15 at 19:28

One I really like is this one :

If $f$ is a continuous function for which $f(a+b-t)=f(t)$ then $$\int_a^b t\cdot f(t) \mathrm{d}t=\frac{a+b}{2}\int_a^bf(t) \mathrm{d}t$$

Example :

$$\begin{align} \int_0^{\pi} \frac{x\sin(x)}{1+\cos^2 (x)}\mathrm{d}x &=\frac{\pi}{2}\int_0^{\pi} \frac{\sin(x)}{1+\cos^2 (x)}\mathrm{d}x\\ &=\frac{\pi}{2} \left[-\arctan(\cos(x))\right]_0^{\pi} \\ &=\frac{\pi^2}{4}\end{align}$$

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