Lesser-known integration tricks

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 that $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$, 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.

Solution 1:

I don't know about "lesser known" but many calculus courses pass over hyperbolic functions. 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.

Solution 2:

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 $$\sin{x}=2\cos{\frac{x}{2}}\sin{\frac{x}{2}}=\frac{2\cos{\frac{x}{2}}\sin{\frac{x}{2}}}{\cos^2{\frac{x}{2}}+\sin^2{\frac{x}{2}}}=\frac{2u}{1+u^2}$$ $$\cos{x}=\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}=\frac{\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}}{\cos^2{\frac{x}{2}}+\sin^2{\frac{x}{2}}}=\frac{1-u^2}{1+u^2}$$

Solution 3:

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.

Solution 4:

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. :-)

Solution 5:

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.