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I am currently studying for the GRE math subject test, which heavily tests calculus. I've reviewed most of the basic calculus techniques (integration by parts, trig substitutions, etc.) I am now looking for a list or reference for some lesser-known tricks or clever substitutions that are useful in integration. For example, I learned of this trick

$\int_a^b f(x) \; dx = \int_a^b f(a + b -x) \; dx$

in the question showing $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$ when $f$ is even

I am especially interested in tricks that can be used without an excessive amount of computation, as I believe (or hope?) that these will be what is useful for the GRE.

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4  
I have to wonder if the Weierstrass substitution counts as "lesser-known"... –  J. M. Oct 9 '11 at 0:32
    
I also don't know if this trick for doubly-infinite integrals is well known: $\int_{-\infty}^\infty f(t)\mathrm dt=\frac12\int_{-\infty}^\infty (f(t)+f(-t))\mathrm dt=\int_0^\infty (f(t)+f(-t))\mathrm dt$. It is usual that the last two integrals are more manageable than the first. –  J. M. Oct 9 '11 at 1:00
    
@J.M. $\int_{-\infty}^{\infty} f(t) dt = \int_0^{\infty} (f(t) + f(-t)) dt$ might not hold for $f(t) = 2t/(1+t^2)$ because the integral on the left is undefined (works out to $\infty - \infty$) while the one on the right is $0$ –  Dilip Sarwate Oct 9 '11 at 1:40
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@Dilip: But the Cauchy principal value of the integral of your function is indeed zero. :) –  J. M. Oct 9 '11 at 1:54
1  
Another one: it is sometimes helpful to express trigonometric/hyperbolic functions in terms of (complex) exponentials; this allows you to readily do things like partial fraction decomposition... –  J. M. Oct 9 '11 at 3:00

7 Answers 7

I don't know about "lesser known" but many calculus courses pass over hyperbolic functions: http://en.wikipedia.org/wiki/Hyperbolic_function

Just as the identity $\sin^2(t)+\cos^2(t)=1$ allows one to deal with $1-x^2$ terms, the identity $\cosh^2(t)-\sinh^2(t)=1$ allows one to deal with $1+x^2$ terms.

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14  
A lot of times, using $\cosh$ and $\sinh$ makes for more manageable integrals than using $\sec$ and $\tan$... –  J. M. Oct 9 '11 at 0:30
    
I haven't even formally learned what hyperbolic trig. functions are! –  zerosofthezeta Dec 25 '13 at 8:42

Maybe for your purposes the Weierstrass substitutiontangent half-angle substitution could be considered "lesser known", although lots of textbooks have it. [PS added on Christmas 2013: Since the time this answer was posted, it's been pointed out that Weierstrass never wrote anything about this substitution, but Euler did, during the century before Weierstrass lived. It is not clear to me that the name "Weierstrass substitution" comes from anywhere besides Stewart's calculus text.]

Still less well known is differentiation under the integral sign.

The GRE math subject test might do some contour integration. Here you'd see integrals that might superficially look as innocent as any you see in first-year calculus but you use complex variables to find them. I remember that when I took the test, there was one question about residues.

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This is not a very deep thing, but it's often convenient to do repeated integrations by parts all in one fell swoop, especially when one factor is a polynomial so that the process terminates after finitely many steps. For example, to compute the sine Fourier series of $f(t) = t^3 + a t^2 + bt + c$ one wants the antiderivative of $f(t) \sin(n\Omega t)$. Easy: $$ \begin{array} {}\int (t^3 + a t^2 + bt + c) \sin(n\Omega t) \;dt =&{}+ (t^3 + a t^2 + bt + c) \frac{-\cos(n\Omega t)}{n\Omega} \\ &{}- (3t^2 + 2a t + b) \frac{-\sin(n\Omega t)}{(n\Omega)^2} \\ &{}+ (6 t + 2a) \frac{\cos(n\Omega t)}{(n\Omega)^3} \\ &{}- 6 \frac{\sin(n\Omega t)}{(n\Omega)^4} \\ &{}+ C. \end{array} $$ Notice the pattern with alternating signs: $$ +,-,+,-,\ldots, $$ successive derivatives of one factor: $$ t^3 + a t^2 + bt + c, \quad 3t^2 + 2a t + b, \quad 6 t + 2a, \quad 6, \quad 0, $$ and successive antiderivatives of the other factor: $$ \sin(n\Omega t), \quad \frac{-\cos(n\Omega t)}{n\Omega}, \quad \frac{-\sin(n\Omega t)}{(n\Omega)^2}, \quad \frac{\cos(n\Omega t)}{(n\Omega)^3}, \quad \frac{\sin(n\Omega t)}{(n\Omega)^4}, \quad \ldots, $$ and the process stops when the derivatives reach zero.

Countless times, I've seen students make sign errors in this type of integral that could have been avoided by organizing the computations according to these simple rules.

Apparently this is being taught as a trick in some schools, judging from this clip from the 1988 movie Stand and Deliver. :-)

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For integrating rational expressions of sine or cosine, the substitution $u=\tan{\frac{x}{2}}$ always leads to a rational function in $u$. We have $$\begin{array}{ll} u=\tan{\frac{x}{2}}, & dx=\frac{2du}{1+u^2} \end{array}$$ and $$\sin{x}=2\cos{\frac{x}{2}}\sin{\frac{x}{2}}=\frac{2\cos{\frac{x}{2}}\sin{\frac{x}{2}}}{\cos^2{\frac{x}{2}}+\sin^2{\frac{x}{2}}}=\frac{2u}{1+u^2}$$ $$\cos{x}=\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}=\frac{\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}}{\cos^2{\frac{x}{2}}+\sin^2{\frac{x}{2}}}=\frac{1-u^2}{1+u^2}$$

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http://en.wikipedia.org/wiki/Euler_substitution

I honestly didn't know this useful method existed until I saw a problem on Yahoo

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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, $f = g + h$. 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.

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When integrating rational functions by partial fractions decomposition, the trickiest type of antiderivative that one might need to compute is $$I_n = \int \frac{dx}{(1+x^2)^n}.$$ (Integrals involving more general quadratic factors can be reduced to such integrals, plus integrals of the much easier type $\int \frac{x \, dx}{(1+x^2)^n}$, with the help of substitutions of the form $x \mapsto x+a$ and $x \mapsto ax$.)

For $n=1$, we know that $I_1 = \int \frac{dx}{1+x^2} = \arctan x + C$, and the usual suggestion for finding $I_n$ for $n \ge 2$ is to work one's way down to $I_1$ using the reduction formula $$ I_n = \frac{1}{2(n-1)} \left( \frac{x}{(1+x^2)^{n-1}} + (2n-3) \, I_{n-1} \right) . $$ However, this formula is not easy to remember, and the computations become quite tedious, so the lesser-known trick that I will describe here is (in my opinion) a much simpler way.

From now on, I will use the abbreviation $$T=1+x^2.$$ First we compute $$ \frac{d}{dx} \left( x \cdot \frac{1}{T^n} \right) = 1 \cdot \frac{1}{T^n} + x \cdot \frac{-n}{T^{n+1}} \cdot 2x = \frac{1}{T^n} - \frac{2n x^2}{T^{n+1}} = \frac{1}{T^n} - \frac{2n (T-1)}{T^{n+1}} \\ = \frac{1}{T^n} - \frac{2n T}{T^{n+1}} + \frac{2n}{T^{n+1}} = \frac{2n}{T^{n+1}} - \frac{2n-1}{T^n} . $$ Let us record this result for future use, in the form of an integral: $$ \int \left( \frac{2n}{T^{n+1}} - \frac{2n-1}{T^n} \right) dx = \frac{x}{T^n} + C . $$ That is, we have $$ \begin{align} \int \left( \frac{2}{T^2} - \frac{1}{T^1} \right) dx &= \frac{x}{T} + C ,\\ \int \left( \frac{4}{T^3} - \frac{3}{T^2} \right) dx &= \frac{x}{T^2} + C ,\\ \int \left( \frac{6}{T^4} - \frac{5}{T^3} \right) dx &= \frac{x}{T^3} + C ,\\ &\vdots \end{align} $$ With the help of this, we can easily compute things like $$ \begin{align} \int \left( \frac{1}{T^3} + \frac{5}{T^2} - \frac{2}{T} \right) dx &= \int \left( \frac14 \left( \frac{4}{T^3} - \frac{3}{T^2} \right) + \frac{\frac34 + 5}{T^2} - \frac{2}{T} \right) dx \\ &= \int \left( \frac14 \left( \frac{4}{T^3} - \frac{3}{T^2} \right) + \frac{23}{4} \cdot \frac12 \left( \frac{2}{T^2} - \frac{1}{T^1} \right) + \frac{\frac{23}{8}-2}{T} \right) dx \\ &= \frac14 \frac{x}{T^2} + \frac{23}{8} \frac{x}{T} + \frac{7}{8} \arctan x + C . \end{align} $$ Of course, the relation that we are using, $\frac{2n}{I_{n+1}}-\frac{2n-1}{I_n}=\frac{x}{T^n}$, really is the reduction formula in disguise. However, the trick is:

(a) to derive the formula just by differentiation (instead of starting with an integral where the exponent is one step lower than the one that we're interested in, inserting a factor of $1$, integrating by parts, and so on),

and

(b) to leave the formula in its "natural" form as it appears when differentiating (instead of solving for $I_{n+1}$ in terms of $I_n$), which results in a structure which is easier to remember and a more pleasant way of organizing the computations.

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