Integral Calculus 201-SN3-RE • CEGEP, Montreal

Revision sheet: integration tricks (201-SN3-RE)

After the four classic techniques, the 201-SN3-RE (formerly 201-NYB) course still holds a handful of tricks that serve exactly once per exam: those that exploit a SYMMETRY, of the domain or of the function. They have one thing in common: they turn the integral into an equation whose unknown is the integral itself.

This sheet gives the five tricks with their conditions of use, and above all the continuity and monotonicity traps that make them wrong when they are applied mechanically. It is written for CEGEP students in the Science program (Sciences de la nature); once it is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

These techniques almost never compute the integral directly: they produce a SECOND equation in $I$ . Looking for a result on the first pass means missing the only thing they know how to do.

This chapter is part of 201-SN3-RE, Integral Calculus

The essentials

Symmetries of the domain

• KING'S RULE: $\displaystyle\int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx$ . It computes nothing by itself; it provides a SECOND equation in $I$ .
• PARITY: on $[-a, a]$ , an ODD function gives $0$ , and an EVEN function gives $2\displaystyle\int_{0}^{a}f$ . Parity must be checked, not assumed.
• PERIODICITY: $\displaystyle\int_{a}^{a+T}f=\int_{0}^{T}f$ for every function of period $T$ , whatever $a$ is.
• SYMMETRY $x\leftrightarrow\dfrac{1}{x}$ : $\displaystyle\int_{1/a}^{a}\frac{g(x)}{x}dx=\int_{1/a}^{a}\frac{g(1/x)}{x}dx$ , the factor $\dfrac{1}{x}$ being what makes the swap possible.
• All these equalities are proved by a simple substitution, and it is that substitution you must be able to write, not the formula.

The two curves are images of each other under the folding of the interval: they therefore bound the SAME area. Here their sum equals

1

everywhere, which gives

2I=\dfrac{\pi}{2}

in one line.

King's rule folds the interval onto itself. It is decisive when $f(x)+f(a+b-x)$ simplifies, typically to a constant, which gives $2I$ on one side and a trivial integral on the other.

Weierstrass and hyperbolic functions

• WEIERSTRASS: with $t=\tan\dfrac{x}{2}$ , we have $\sin x=\dfrac{2t}{1+t^{2}}$ , $\cos x=\dfrac{1-t^{2}}{1+t^{2}}$ and $dx=\dfrac{2\,dt}{1+t^{2}}$ .
• It rationalizes EVERY fraction in $\sin$ and $\cos$ , but it is only valid on an interval where $\tan\dfrac{x}{2}$ is CONTINUOUS, so never across $x=\pi$ .
• HYPERBOLIC FUNCTIONS: $\cosh^{2}u-\sinh^{2}u=1$ , with a MINUS sign, and $(\sinh u)'=\cosh u$ , $(\cosh u)'=\sinh u$ , with no minus sign.
• For $\sqrt{x^{2}+a^{2}}$ , set $x=a\sinh u$ ; for $\sqrt{x^{2}-a^{2}}$ , set $x=a\cosh u$ . The antiderivatives obtained are often simpler than with trigonometric substitution.
• Logarithmic forms: $\sinh^{-1}x=\ln\left(x+\sqrt{x^{2}+1}\right)$ and $\cosh^{-1}x=\ln\left(x+\sqrt{x^{2}-1}\right)$ .

Where

t=\tan\frac{x}{2}

comes from: the line from

(-1, 0)

to the point

M

at angle

x

makes the angle

\frac{x}{2}

with the horizontal axis (inscribed angle theorem), so its slope is

t

Weierstrass is the technique of last resort: it always works but often produces a heavy rational function. Use it only after looking for a shorter trigonometric identity.

Reduction formulas and Wallis integrals

• A reduction formula expresses $I_{n}$ in terms of $I_{n-2}$ : it is obtained by an integration by parts carried out with $n$ kept symbolic.
• WALLIS: $I_{n}=\displaystyle\int_{0}^{\pi/2}\sin^{n}x\,dx$ satisfies $I_{n}=\dfrac{n-1}{n}I_{n-2}$ , and the same formula holds for $\cos^{n}x$ .
• BOTH base cases are essential: $I_{0}=\dfrac{\pi}{2}$ and $I_{1}=1$ . An even $n$ goes down to $I_{0}$ and brings in $\pi$ ; an odd $n$ goes down to $I_{1}$ and never brings one in.
• A useful relation: $I_{n}I_{n-1}=\dfrac{\pi}{2n}$ , which serves as an immediate check.
• System of two integrals: when $I$ and $J$ appear together, write $I+J$ and $I-J$ , both computable, then solve.

The curves of

\sin^{n}x

(n = 1, 2, 4, 8) close in on

\frac{\pi}{2}

n

grows: the area

I_{n}

decreases to

0

, which is what

I_{n}=\frac{n-1}{n}I_{n-2}

says.

A $\pi$ in the result of a Wallis integral with an ODD exponent is always an error: the descent stops at $I_{1}=1$ , and no $\pi$ can appear.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Expecting King's rule to compute the integral

the whole question, 6 points

What not to write

“I apply King's rule, I get $\displaystyle\int_{0}^{\pi/2}\frac{\cos x}{\sin x+\cos x}dx$ , and I am no further ahead.”

What to write

“That is exactly the goal: this second integral $J$ equals $I$ , and $I+J=\displaystyle\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2}$ , so $I=\frac{\pi}{4}$ .”

Why: The trick produces an EQUATION, never a result. The only question to ask next is: does the sum of the two integrands simplify? If it does, the problem is over.

2. Applying Weierstrass across a discontinuity

the whole question, 6 points, with a zero result for a strictly positive integrand

What not to write

“ $\displaystyle\int_{0}^{2\pi}\frac{dx}{2+\cos x}$ : with $t=\tan\frac{x}{2}$ , the bounds become $t=0$ and $t=0$ , so the integral equals $0$ .”

What to write

“ $\tan\dfrac{x}{2}$ blows up at $x=\pi$ , in the MIDDLE of the interval: the substitution is forbidden. Split into $[0, \pi]$ and $[\pi, 2\pi]$ , or use symmetry. The exact value is $\dfrac{2\pi}{\sqrt{3}}\approx 3.63$ .”

The substitution starts from

t=0

and comes back to it, but it goes through infinity in the middle of the interval. An integral between two equal bounds is zero, which is absurd for a positive integrand.

Why: The integrand is everywhere greater than $\dfrac{1}{3}$ : its integral over an interval of length $2\pi$ cannot be zero. That check exposes the error before any reasoning.

3. Invoking parity without checking that the function is odd

the whole question, 4 points, with a zero result for a positive integrand

What not to write

“The interval is symmetric, so $\displaystyle\int_{-2}^{2}x^{2}\,dx=0$ .”

What to write

“ $x^{2}$ is EVEN: the integral equals $2\displaystyle\int_{0}^{2}x^{2}dx=\dfrac{16}{3}$ . Only an ODD function vanishes over a symmetric interval.”

Here the two lobes have the same area and opposite signs: the integral is indeed zero. With

x^{2}

, both lobes would lie on the SAME side of the axis and nothing would cancel.

Why: The symmetry of the INTERVAL is not enough: it is the symmetry of the FUNCTION that decides. A positive integrand can never give zero, whatever the interval.

4. Starting the Wallis recurrence without its base cases

3 points, and a result without

\pi

where there should be one

What not to write

“ $I_{6}=\dfrac{5}{6}\cdot\dfrac{3}{4}\cdot\dfrac{1}{2}\cdot I_{0}$ , and I take $I_{0}=1$ .”

What to write

“ $I_{0}=\displaystyle\int_{0}^{\pi/2}dx=\dfrac{\pi}{2}$ , not $1$ . So $I_{6}=\dfrac{15}{48}\cdot\dfrac{\pi}{2}=\dfrac{5\pi}{32}$ . It is $I_{1}$ that equals $1$ .”

Why: The two base cases do not look alike: $I_{0}$ integrates the constant function $1$ over a quarter turn, hence the $\pi$ ; $I_{1}$ integrates a sine, hence exactly $1$ .

5. Forgetting the Jacobian in the substitution $x\mapsto\dfrac{1}{x}$

the whole question, 5 points

What not to write

“I set $x=\dfrac{1}{u}$ , so $\displaystyle\int f(x)\,dx=\int f\!\left(\frac{1}{u}\right)du$ .”

What to write

“ $dx=-\dfrac{du}{u^{2}}$ : the factor $\dfrac{1}{u^{2}}$ and the minus sign, which swaps the bounds, are part of the substitution.”

Why: It is precisely this factor $\dfrac{1}{u^{2}}$ that makes the trick work: combined with a $\dfrac{1}{x}$ already present in the integrand, it rebuilds a $\dfrac{1}{u}$ and closes the symmetry.

6. Confusing the hyperbolic identity with the trigonometric one

the whole question, 5 points

What not to write

“ $\cosh^{2}u+\sinh^{2}u=1$ , as for sine and cosine.”

What to write

“ $\cosh^{2}u-\sinh^{2}u=1$ , with a MINUS sign. On the other hand, the derivatives have NO minus sign: $(\cosh u)'=\sinh u$ .”

The circle satisfies

\cos^{2}t+\sin^{2}t=1

, the hyperbola

\cosh^{2}u-\sinh^{2}u=1

: a PLUS for one, a MINUS for the other.

Why: The signs are exactly reversed compared with the trigonometric case: minus in the identity, plus in the derivatives. The check is immediate: $\cosh u\ge 1$ always, so the sum of the squares cannot equal $1$ .

7. Treating an improper integral like an ordinary one

the whole question, 5 points

What not to write

“ $\displaystyle\int_{-1}^{1}\frac{dx}{x^{2}}=\left[-\frac{1}{x}\right]_{-1}^{1}=-1-1=-2$ .”

What to write

“The integrand is not defined at $0$ , which is INSIDE the interval: the integral is improper and it DIVERGES. The value $-2$ is doubly absurd, since the integrand is positive.”

\frac{1}{x^{2}}

is positive and blows up at

0

, INSIDE the interval: the area is infinite, and the

-2

of the blind computation is impossible for a positive function.

Why: The fundamental theorem requires continuity on the whole interval. Applying the antiderivative across a discontinuity regularly produces negative results for positive functions.

8. Using Weierstrass when an identity is enough

no points, but ten minutes on the exam

What not to write

“ $\displaystyle\int\frac{\sin x}{1+\cos x}dx$ : I set $t=\tan\frac{x}{2}$ and rationalize.”

What to write

“The derivative of $1+\cos x$ is $-\sin x$ , which is already there: $u=1+\cos x$ gives $-\ln\left|1+\cos x\right|+C$ in two lines.”

Why: Weierstrass rationalizes everything, but produces a heavy fraction. Bring it out only after looking for a derivative already present and a double-angle identity.

9. Writing the system of two integrals with the wrong sign

2 points, and the wrong one of the two answers

What not to write

“ $I+J$ and $J-I$ are both computable, so $I=\dfrac{(I+J)+(J-I)}{2}$ .”

What to write

“ $I=\dfrac{(I+J)-(J-I)}{2}$ . The right-hand side of the wrong computation equals $J$ , not $I$ : it is the other integral you would have obtained.”

Why: The system in two unknowns is solved like any linear system. The error gives a plausible result, sometimes even a correct one if $I=J$ , which makes it hard to spot.

10. Applying a reduction formula without checking the parity of $n$

3 points

What not to write

“ $I_{5}=\dfrac{4}{5}\cdot\dfrac{2}{3}\cdot I_{1}$ , and I replace $I_{1}$ with $\dfrac{\pi}{2}$ .”

What to write

“The descent from an ODD $n$ stops at $I_{1}=1$ : $I_{5}=\dfrac{8}{15}$ , without any $\pi$ .”

Why: The parity of $n$ decides where the recurrence ends, hence whether $\pi$ appears in the result. It is the fastest check on any Wallis integral.

Which method to choose

Which trick to try, and in which order

The shape of the integral, once the four classic techniques have failed

If the interval is symmetric about $0$ → test the PARITY of the integrand: odd gives $0$ , even gives twice the half

Example: $\displaystyle\int_{-2}^{2}x^{3}dx=0$
If the integrand shows a symmetry between $x$ and $a+b-x$ → KING'S RULE, then add the two expressions

Example: $\displaystyle\int_{0}^{\pi/2}\frac{\sin x}{\sin x+\cos x}dx=\frac{\pi}{4}$
If the integrand is a FRACTION in $\sin$ and $\cos$ that resists → WEIERSTRASS, after checking that $\tan\dfrac{x}{2}$ is continuous on the interval

Example: $\displaystyle\int_{0}^{\pi/2}\frac{dx}{2+\cos x}$
If the exponent is a generic integer $n$ → REDUCTION FORMULA by parts, keeping $n$ symbolic

Example: $I_{n}=\dfrac{n-1}{n}I_{n-2}$ for the Wallis integrals
If the bounds are $\dfrac{1}{a}$ and $a$ , or $0$ and infinity → substitution $x\mapsto\dfrac{1}{x}$ , without forgetting $dx=-\dfrac{du}{u^{2}}$

it closes the integral on itself when a factor $\dfrac{1}{x}$ is present
If a root $\sqrt{x^{2}\pm a^{2}}$ resists trigonometry → HYPERBOLIC substitution, often shorter and with no absolute value

$x=a\sinh u$ for a plus, $x=a\cosh u$ for a minus

The order matters: parity costs five seconds, King's rule one line, Weierstrass one page. So try them in that order, not in the order of the course.

Before using a trick, check five things

The conditions of validity, to check BEFORE the computation

If you substitute → check that the substitution is MONOTONIC and CONTINUOUS on the whole interval

Example: $\tan\dfrac{x}{2}$ is forbidden on $[0, 2\pi]$
If the integrand has a denominator → check that it does NOT vanish in the interval, endpoints included

Example: $\displaystyle\int_{-1}^{1}\frac{dx}{x^{2}}$ diverges
If you invoke parity → check oddness through $f(-x)=-f(x)$ , not just the symmetry of the interval

Example: $x^{2}$ is even, its symmetric integral does not vanish
If you use a recurrence → identify the base cases AND the parity of the index, which decides where it ends

Example: $I_{0}=\dfrac{\pi}{2}$ but $I_{1}=1$
If the result is negative → check the sign of the integrand: a positive function cannot give a negative integral

it is the check that catches almost every error in this chapter

These five checks take less than a minute and save half of the marks lost. The tricks are powerful precisely because they short-circuit the computation, so they also short-circuit the safeguards.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up King's rule

When to use it: The integral is definite, its bounds are symmetric about a centre, and the integrand shows a symmetry between $x$ and $a+b-x$ .

1 Name the integral $I$ and write it out, so that you can then treat it as an unknown.
2 Apply the substitution $u=a+b-x$ and check that it is decreasing, hence that it swaps the bounds: that is what restores the direction of the integral.
3 Write the second expression obtained, naming it $J$ if it differs from $I$ .
4 Add the two expressions and check that the sum of the integrands simplifies; if it does not, the trick is useless and you must change method.
5 Solve the resulting equation in $I$ and conclude.

Concluding sentence

“Setting $u=\dfrac{\pi}{2}-x$ , the integral $I$ becomes $J=\displaystyle\int_{0}^{\pi/2}\frac{\cos u}{\sin u+\cos u}du$ ; now $I+J=\displaystyle\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2}$ and $I=J$ by symmetry, so $I=\frac{\pi}{4}$ .”

The trap: Forgetting to justify that $I=J$ . It is not automatic: it holds here because the substitution literally turns $I$ into $J$ , which proves they are equal. Without that sentence, the equation $I+J=\dfrac{\pi}{2}$ is not enough to conclude.

Marking: Typically 1 point for naming $I$ , 2 for the substitution, 2 for the justified equality $I=J$ , 2 for the sum and its simplification, 1 for the conclusion.

Establishing and applying a reduction formula

When to use it: The integrand carries a generic integer exponent $n$ , or the question asks for a value at a large $n$ .

1 Write $I_{n}$ with the symbolic exponent, and split off a factor to prepare an integration by parts.
2 Integrate by parts keeping $n$ in the computations, without ever giving it a value.
3 Make $I_{n}$ or $I_{n-2}$ reappear on the right-hand side, then isolate $I_{n}$ .
4 Identify the base cases, distinguishing an even index from an odd one.
5 Go down the recurrence to the base case matching the parity of $n$ , and conclude.

Concluding sentence

“The formula $I_{n}=\dfrac{n-1}{n}I_{n-2}$ gives, for $n=6$ even, $I_{6}=\dfrac{5}{6}\cdot\dfrac{3}{4}\cdot\dfrac{1}{2}\cdot I_{0}=\dfrac{15}{48}\cdot\dfrac{\pi}{2}=\dfrac{5\pi}{32}$ .”

The trap: Going down to $I_{0}$ for an odd index. The recurrence jumps two at a time: an odd index never meets $I_{0}$ and stops at $I_{1}=1$ .

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The sign of the result against the sign of the integrand

A positive integrand on the interval always gives a positive integral. It is the check that catches almost every error in this chapter.

A result of $-2$ for $\displaystyle\int_{-1}^{1}\frac{dx}{x^{2}}$ : the integral is in fact divergent.
Continuity on the whole interval

Before any substitution or evaluation, locate the points where the integrand or the substitution blows up, and check that they lie outside the interval.

A substitution by $\tan\dfrac{x}{2}$ on $[0, 2\pi]$ : the point $\pi$ is inside the interval.
Whether $\pi$ appears in a Wallis integral

An EVEN index gives a result containing $\pi$ ; an ODD index gives a pure rational number. The check is instantaneous.

An $I_{5}$ containing $\pi$ : the descent stopped at the wrong base case.
The relation $I_{n}I_{n-1}=\dfrac{\pi}{2n}$

Multiply two consecutive Wallis terms: the product must equal $\dfrac{\pi}{2n}$ . This validates two values at once.

$I_{6}I_{5}=\dfrac{5\pi}{32}\cdot\dfrac{8}{15}=\dfrac{\pi}{12}$ , as expected.
The order of magnitude by bounding

Bound the integrand between two constants on the interval and multiply by the length: the result must fall within those bounds.

$\dfrac{1}{2+\cos x}$ lies between $\dfrac{1}{3}$ and $1$ , so its integral over $[0, 2\pi]$ lies between $2.09$ and $6.28$ : the value $3.63$ fits, the value $0$ does not.

The typical problem, taken apart

An integral folded onto itself by King's rule

Compute $I=\displaystyle\int_{0}^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx$ .

Then apply the same method to recover $\displaystyle\int_{0}^{\pi/2}\sin^{6}x\,dx$ with the reduction formula, and check it with the Wallis relation.

Step 1

Name $I=\displaystyle\int_{0}^{\pi/2}\frac{\sin x}{\sin x+\cos x}dx$ . None of the four classic techniques applies: the denominator is neither a derivative already present, nor a rational function, nor a product.

Why

Naming the integral is the first step, because the method consists precisely in treating it as an UNKNOWN. Noting that the ordinary methods fail justifies moving on to the tricks.

Step 2

King's rule: set $u=\dfrac{\pi}{2}-x$ , so $du=-dx$ ; the bounds swap, which compensates for the sign. We get $I=\displaystyle\int_{0}^{\pi/2}\frac{\sin\left(\frac{\pi}{2}-u\right)}{\sin\left(\frac{\pi}{2}-u\right)+\cos\left(\frac{\pi}{2}-u\right)}du$ .

Why

The substitution is decreasing: it swaps the bounds, and that second swap cancels the minus sign of $du$ . It is this double effect that makes the identity hold as written.

Step 3

The identities $\sin\left(\dfrac{\pi}{2}-u\right)=\cos u$ and $\cos\left(\dfrac{\pi}{2}-u\right)=\sin u$ give $I=\displaystyle\int_{0}^{\pi/2}\frac{\cos u}{\cos u+\sin u}du$ , which we call $J$ .

Why

The integral $J$ is indeed the expected second equation. The crucial point is that the substitution TRANSFORMED $I$ into $J$ , which proves $I=J$ with no further argument.

Step 4

Sum: $I+J=\displaystyle\int_{0}^{\pi/2}\frac{\sin x+\cos x}{\sin x+\cos x}dx=\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2}$ .

Why

This is where the trick pays off: the sum of the two integrands simplifies to the constant $1$ . Had it not simplified, the trick would have been useless.

Step 5

Since $I=J$ , we have $2I=\dfrac{\pi}{2}$ , so $I=\dfrac{\pi}{4}\approx 0.785$ .

Why

The equation in $I$ is solved in one line. It is the pattern shared by every trick in this chapter: produce an equation rather than a computation.

Step 6

Check: the integrand lies between $0$ and $1$ on $\left[0, \dfrac{\pi}{2}\right]$ , so $I$ lies between $0$ and $\dfrac{\pi}{2}\approx 1.57$ . The value $0.785$ is exactly half of it, which is consistent with the symmetry.

Why

The bounding is done mentally and confirms the order of magnitude. The position in the middle of the bounds is in fact a direct consequence of $I=J$ .

Step 7

Second part: $I_{6}=\displaystyle\int_{0}^{\pi/2}\sin^{6}x\,dx$ . The reduction formula gives $I_{6}=\dfrac{5}{6}I_{4}=\dfrac{5}{6}\cdot\dfrac{3}{4}I_{2}=\dfrac{5}{6}\cdot\dfrac{3}{4}\cdot\dfrac{1}{2}I_{0}$ .

Why

The index $6$ is EVEN: the descent stops at $I_{0}$ , and that is where the $\pi$ will come from. An odd index would have stopped at $I_{1}=1$ , without $\pi$ .

Step 8

With $I_{0}=\dfrac{\pi}{2}$ : $I_{6}=\dfrac{15}{48}\cdot\dfrac{\pi}{2}=\dfrac{5\pi}{32}\approx 0.491$ .

Why

The value of $I_{0}$ is the only place where an error is possible in this descent: the integral of the constant function $1$ over a quarter turn is indeed $\dfrac{\pi}{2}$ .

Step 9

Check with the Wallis relation: $I_{5}=\dfrac{4}{5}\cdot\dfrac{2}{3}\cdot 1=\dfrac{8}{15}$ , and $I_{6}I_{5}=\dfrac{5\pi}{32}\cdot\dfrac{8}{15}=\dfrac{\pi}{12}$ , exactly $\dfrac{\pi}{2\times 6}$ .

Why

This relation validates TWO consecutive values with a single computation. In particular, it catches the base-case error, which would immediately break the product.

The conclusion, written out

“ $\displaystyle\int_{0}^{\pi/2}\frac{\sin x}{\sin x+\cos x}dx=\frac{\pi}{4}$ , obtained by folding the interval onto itself; and $\displaystyle\int_{0}^{\pi/2}\sin^{6}x\,dx=\frac{5\pi}{32}$ , obtained by taking the reduction formula down to $I_{0}=\frac{\pi}{2}$ .”

The classic mistake on this problem: Looking for an antiderivative of $\dfrac{\sin x}{\sin x+\cos x}$ . It exists and reads $\dfrac{x}{2}-\dfrac{1}{2}\ln\left|\sin x+\cos x\right|$ , but getting it takes a much longer manipulation than the three lines of King's rule.

Learn by heart

• King's rule: $\displaystyle\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ . It gives an EQUATION, not a result.
• Parity: only an ODD function vanishes over a symmetric interval.
• Weierstrass: $t=\tan\dfrac{x}{2}$ , with $\sin x=\dfrac{2t}{1+t^{2}}$ , $\cos x=\dfrac{1-t^{2}}{1+t^{2}}$ , $dx=\dfrac{2dt}{1+t^{2}}$ ; NEVER across $x=\pi$ .
• Hyperbolic functions: $\cosh^{2}u-\sinh^{2}u=1$ (minus sign), but $(\cosh u)'=\sinh u$ (no minus sign).
• Wallis: $I_{n}=\dfrac{n-1}{n}I_{n-2}$ , with $I_{0}=\dfrac{\pi}{2}$ and $I_{1}=1$ . Even index: some $\pi$ ; odd index: no $\pi$ at all.
• Wallis check: $I_{n}I_{n-1}=\dfrac{\pi}{2n}$ .
• Substitution $x\mapsto\dfrac{1}{x}$ : never forget $dx=-\dfrac{du}{u^{2}}$ .
• A negative result for a positive integrand signals a discontinuity or a substitution that is not monotonic.

Frequently asked questions

What is King's rule for, if it does not compute the integral?

It provides a second expression of the same integral. Adding the two often gives an integrand that simplifies, typically to a constant. You are then left with an equation whose unknown is the integral, and solving it takes one line. It is a method of equations, not a method of computation.

When is the Weierstrass substitution forbidden?

As soon as the interval of integration contains a point where the tangent of half the angle is undefined, that is, pi and its translates by multiples of two pi. On such an interval, the substitution goes through infinity and the two bounds may even give the same value, which produces an absurd zero integral.

Is an integral over a symmetric interval always zero?

No, only if the function is odd. An even function, on the contrary, gives twice its integral over half the interval. The symmetry of the interval is necessary but not sufficient: you must check the relation between the value at minus x and the value at x.

What are the base cases of the Wallis integrals?

The integral of index zero equals pi over two, because it integrates the constant function one over a quarter turn. The integral of index one equals exactly one. The recurrence goes down two at a time, so an even index ends at the first case and brings in pi, while an odd index ends at the second and gives a rational number.

Why use hyperbolic functions rather than trigonometry?

Because they often avoid absolute values and painful returns to the original variable. For the square root of x squared plus a squared, the hyperbolic sine substitution leads straight to a logarithm, whereas the trigonometric tangent goes through a secant and a triangle. Watch out for the minus sign in the fundamental identity.

How can I quickly spot an error in this chapter?

By looking at the sign. These tricks short-circuit the computation, so they also short-circuit the safeguards. A positive integrand must give a positive integral, and a zero integral over an interval where the function does not change sign is impossible. It is the check that catches almost every error.

Practise it

Corrected exercises: Integration tricks: Weierstrass, King's rule, Wallis, 201-SN3-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc2-integration-tricks. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: integration tricks (201-SN3-RE)

The essentials

Symmetries of the domain

Weierstrass and hyperbolic functions

Reduction formulas and Wallis integrals

The mistakes that cost marks

1. Expecting King's rule to compute the integral

2. Applying Weierstrass across a discontinuity

3. Invoking parity without checking that the function is odd

4. Starting the Wallis recurrence without its base cases

5. Forgetting the Jacobian in the substitution x↦1xx\mapsto\dfrac{1}{x}x↦x1​

6. Confusing the hyperbolic identity with the trigonometric one

7. Treating an improper integral like an ordinary one

8. Using Weierstrass when an identity is enough

9. Writing the system of two integrals with the wrong sign

10. Applying a reduction formula without checking the parity of nnn

Which method to choose

Which trick to try, and in which order

Before using a trick, check five things

How the answer is expected to be written

Writing up King's rule

Establishing and applying a reduction formula

Check before you hand in

The typical problem, taken apart

An integral folded onto itself by King's rule

Learn by heart

Frequently asked questions

What is King's rule for, if it does not compute the integral?

When is the Weierstrass substitution forbidden?

Is an integral over a symmetric interval always zero?

What are the base cases of the Wallis integrals?

Why use hyperbolic functions rather than trigonometry?

How can I quickly spot an error in this chapter?

Corrected exercises: Integration tricks: Weierstrass, King's rule, Wallis, 201-SN3-RE

See also

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5. Forgetting the Jacobian in the substitution $x\mapsto\dfrac{1}{x}$

10. Applying a reduction formula without checking the parity of $n$