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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 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

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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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- Integral $\int_0^{\infty} \frac{\ln \cos^2 x}{x^2}dx=-\pi$
- How to prove that $b^{x+y} = b^x b^y$ using this approach?

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.

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

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

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

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*. 🙂

Maybe for your purposes the ~~Weierstrass substitution~~tangent 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.

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}

$$

\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.

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, $2n \, I_{n+1} – (2n-1) \, I_n = \frac{x}{T^n}$, 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.

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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