# 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

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

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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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
@Dilip: But the Cauchy principal value of the integral of your function is indeed zero. :) –  Ｊ. Ｍ. Oct 9 '11 at 1:54
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

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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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

Maybe for your purposes the Weierstrass substitution could be considered "lesser known", although lots of textbooks have it.

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