Integral Calculus 201-SN3-RE • CEGEP, Montreal

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

This sheet does not copy out the table of formulas again: you have it in your notes. It deals with what decides the grade on the second test of Integral Calculus 201-SN3-RE (formerly 201-NYB), that is, the CHOICE of technique, made in ten seconds from the form of the integrand, and the three or four moves that make a well-chosen technique fail anyway.

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet 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

You do not pick a technique at random: you read the FORM of the integrand, and each form has its own technique. The time lost on the exam almost always comes from launching an integration by parts where a ten-second substitution would have done the job.

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

The essentials

The three trigonometric substitutions, and their triangle

• $\sqrt{a^{2}-x^{2}}$ : set $x=a\sin\theta$ , which gives $\sqrt{a^{2}-x^{2}}=a\cos\theta$ and $dx=a\cos\theta\,d\theta$ .
• $\sqrt{a^{2}+x^{2}}$ : set $x=a\tan\theta$ , which gives $\sqrt{a^{2}+x^{2}}=a\sec\theta$ and $dx=a\sec^{2}\theta\,d\theta$ .
• $\sqrt{x^{2}-a^{2}}$ : set $x=a\sec\theta$ , which gives $\sqrt{x^{2}-a^{2}}=a\tan\theta$ and $dx=a\sec\theta\tan\theta\,d\theta$ .
• You go back to $x$ through the reference TRIANGLE, never by forcing $\theta=\arcsin\frac{x}{a}$ into the answer.

The three reference triangles: the root is always a side, and the angle

\theta

is read off the triangle when you go back to

x

A radicand that is not already in one of the three forms gets its square completed first: $x^{2}+4x+13=(x+2)^{2}+9$ .

Integration by parts, and the choice of $u$

• $\int u\,dv=uv-\int v\,du$ : you differentiate $u$ , you integrate $dv$ .
• LIATE gives the order of priority for $u$ : Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential.
• The right $u$ is the one that SIMPLIFIES when differentiated; the right $dv$ is the one you know how to integrate.
• $\int\ln x\,dx$ and $\int\arctan x\,dx$ are done by parts with $dv=dx$ : it is the only way.
• Two successive integrations by parts that bring back the original integral: treat it as an unknown and solve for it.

The formula read as areas: the orange area

\int u\,dv

and the blue area

\int v\,du

fill the large rectangle minus the small one,

u_{2}v_{2}-u_{1}v_{1}

If the new integral is MORE complicated than the old one, the choice of $u$ was reversed: start again rather than persevere.

Partial fractions, in order

• First check that the degree of the numerator is STRICTLY less than that of the denominator; otherwise, DIVIDE.
• Factor the denominator completely, over $\mathbb{R}$ .
• Simple linear factor $(x-a)$ : one term $\frac{A}{x-a}$ .
• Repeated linear factor $(x-a)^{2}$ : TWO terms, $\frac{A}{x-a}+\frac{B}{(x-a)^{2}}$ .
• Irreducible quadratic factor $x^{2}+bx+c$ : one term $\frac{Ax+B}{x^{2}+bx+c}$ , with a numerator of degree one.

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. Launching an integration by parts where a substitution is enough

ten minutes, and in the end the whole question

What not to write

“ $\int x\,e^{x^{2}}\,dx$ : I set $u=x$ and $dv=e^{x^{2}}\,dx$ .”

What to write

“ $u=x^{2}$ gives $du=2x\,dx$ , so the integral equals $\frac{e^{x^{2}}}{2}+C$ in one line.”

Why: $e^{x^{2}}$ has no elementary antiderivative: the chosen $dv$ is impossible to integrate, and the solution gets stuck. The right reflex is to test substitution BEFORE any other technique, because it is the only one you can check in ten seconds.

2. Swapping the roles of $u$ and $dv$

the whole question

What not to write

“ $\int x\ln x\,dx$ : I set $u=x$ and $dv=\ln x\,dx$ .”

What to write

“ $u=\ln x$ and $dv=x\,dx$ : the logarithm SIMPLIFIES when differentiated, and $x\,dx$ integrates trivially.”

Why: LIATE puts the logarithm before the algebraic term precisely because it does not integrate easily but differentiates into $\frac{1}{x}$ . The alarm bell rings at once: if the new integral is worse than the old one, the choice was reversed.

3. Losing the sign in the integration by parts formula

the whole question, since the answer is off by an entire term

What not to write

$\int u\,dv=uv+\int v\,du$

What to write

$\int u\,dv=uv-\int v\,du$

Why: The formula comes from integrating $(uv)'=u'v+uv'$ : $uv=\int u'v+\int uv'$ , so $\int uv'=uv-\int u'v$ . The minus sign comes from moving a term to the other side. The check: on $\int x\,e^{x}\,dx$ , the answer is $xe^{x}-e^{x}+C$ , whose derivative does give back $xe^{x}$ .

4. Decomposing into partial fractions without dividing first

the whole question, since the system in

A

is impossible

What not to write

$\frac{x^{2}}{x+1}=\frac{A}{x+1}$

What to write

The degree of the top is not less than that of the bottom: DIVIDE first, $\frac{x^{2}}{x+1}=x-1+\frac{1}{x+1}$ .

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

follows the line

y=x-1

: only the division brings out this polynomial part. A term

\frac{A}{x+1}

on its own tends to

0

and cannot follow it.

Why: Partial fraction decomposition assumes a PROPER rational function. Without division, the system for the coefficients has no solution, and the solution goes round in circles looking for an arithmetic mistake that does not exist.

5. Forgetting a term for a repeated factor

the whole question

What not to write

$\frac{1}{x(x-2)^{2}}=\frac{A}{x}+\frac{B}{(x-2)^{2}}$

What to write

$\frac{1}{x(x-2)^{2}}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^{2}}$

Why: A factor repeated to the power $n$ generates $n$ terms, one for each power from $1$ to $n$ . With only two fractions, the system is overdetermined and has no solution: the dead end comes from the form you set up, not from the computation.

6. Confusing two integrands that differ only by an $x$

the whole question

What not to write

$\int\frac{x\,dx}{x^{2}+1}=\arctan\left(x^{2}+1\right)$

What to write

$\int\frac{x\,dx}{x^{2}+1}=\frac{1}{2}\ln\left(x^{2}+1\right)+C$ , whereas $\int\frac{dx}{x^{2}+1}=\arctan x+C$

Two almost identical integrands:

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

stays positive and gives an arctangent,

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

changes sign at

0

and gives a logarithm.

Why: An $x$ in the numerator turns a form $\frac{1}{1+u^{2}}$ into a form $\frac{u'}{u}$ : two techniques, two antiderivatives with nothing in common. It is the trickiest pair of the chapter, and you must recognize it at first glance.

7. Going back to $x$ without drawing the triangle

1 to 2 marks, since the answer stays unreadable

What not to write

“I found $\theta+\sin\theta\cos\theta+C$ , I replace $\theta$ with $\arcsin\frac{x}{3}$ and I stop there.”

What to write

“The triangle gives $\sin\theta=\frac{x}{3}$ and $\cos\theta=\frac{\sqrt{9-x^{2}}}{3}$ , so the answer is $\arcsin\frac{x}{3}+\frac{x\sqrt{9-x^{2}}}{9}+C$ .”

Why: Leaving a $\sin\left(\arcsin\frac{x}{3}\right)$ in an answer is correct but unusable, and the marking scheme penalizes it. The reference triangle translates each function of $\theta$ into an algebraic expression in $x$ , in three seconds and without any identity to remember.

8. Not completing the square before substituting

the whole question

What not to write

“ $\int\frac{dx}{\sqrt{x^{2}+4x+13}}$ : none of the three forms fits.”

What to write

“ $x^{2}+4x+13=(x+2)^{2}+9$ , so I set $x+2=3\tan\theta$ : it is the form $\sqrt{a^{2}+u^{2}}$ .”

x^{2}+4x+13=(x+2)^{2}+9

: the parabola has its vertex at

(-2, 9)

and never crosses the axis, so the quadratic is irreducible and calls for

x+2=3\tan\theta

Why: The three forms are about $a^{2}\pm u^{2}$ , where $u$ can be any linear expression. Completing the square is the step that reveals this form, and forgetting it leads you to conclude, wrongly, that no technique applies.

Which method to choose

Which technique, according to the form of the integrand

Apply this test in order, and stop at the first branch that applies. Substitution always comes first because it can be tested in ten seconds.

If an inner block and its derivative are both present → SUBSTITUTION, to be tested before anything else

Example: $\int x\,e^{x^{2}}\,dx=\frac{e^{x^{2}}}{2}+C$
If a product of two functions of different kinds, one of which simplifies when differentiated → INTEGRATION BY PARTS, with $u$ chosen according to LIATE

Example: $\int x\ln x\,dx$ with $u=\ln x$
If a quotient of two polynomials → DIVISION if the top is not of strictly lower degree, then PARTIAL FRACTIONS

Example: $\int\frac{x^{2}}{x+1}\,dx=\int\left(x-1+\frac{1}{x+1}\right)dx$
If a root of the form $\sqrt{a^{2}\pm u^{2}}$ or $\sqrt{u^{2}-a^{2}}$ , without the inner derivative → TRIGONOMETRIC SUBSTITUTION, after completing the square if necessary

Example: $\int\frac{dx}{\sqrt{9-x^{2}}}$ with $x=3\sin\theta$
If powers of sine and cosine → an ODD power: split off one factor and substitute; all EVEN: lower the powers with the double-angle formulas

Example: $\int\sin^{3}x\,dx=\int\left(1-\cos^{2}x\right)\sin x\,dx$
If the integrand contains $\ln x$ , $\arcsin x$ or $\arctan x$ ON ITS OWN → integration by parts with $dv=dx$ , there is no other way

Example: $\int\ln x\,dx=x\ln x-x+C$

Two techniques sometimes give the same integral in two forms that differ by a constant: both are correct, and the answer key shows only one.

Which $u$ in an integration by parts

LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Take as $u$ the first item of the list present in the integrand.

If a logarithm is present → $u=\ln$ , always, because it does not integrate simply

Example: $\int x^{2}\ln x\,dx$ with $u=\ln x$
If an arcsine or an arctangent is present → $u$ is that inverse function

Example: $\int\arctan x\,dx$ with $u=\arctan x$ and $dv=dx$
If a polynomial multiplies an exponential or a trigonometric function → $u$ is the POLYNOMIAL, because differentiating lowers its degree by one

Example: $\int x^{2}e^{x}\,dx$ requires two successive integrations by parts
If an exponential multiplies a sine or a cosine → the choice does not matter, but you must KEEP the same one in the second application, then solve for the integral

Example: $\int e^{x}\sin x\,dx=\frac{e^{x}\left(\sin x-\cos x\right)}{2}+C$

The practical rule that sums up LIATE: $u$ is what SIMPLIFIES when differentiated, $dv$ is what lets itself be integrated. If the two conditions contradict each other, the integral calls for another technique.

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.

A trigonometric substitution, written up for the marker

When to use it: Any root of the form $\sqrt{a^{2}-u^{2}}$ , $\sqrt{a^{2}+u^{2}}$ or $\sqrt{u^{2}-a^{2}}$ with no inner derivative available.

1 Identify the form and announce the substitution: “Since $\sqrt{9-x^{2}}$ has the form $\sqrt{a^{2}-x^{2}}$ with $a=3$ , let $x=3\sin\theta$ .”
2 Write the differential and the transformed root: “ $dx=3\cos\theta\,d\theta$ and $\sqrt{9-x^{2}}=3\cos\theta$ .”
3 Rewrite the integral entirely in terms of $\theta$ and compute it.
4 DRAW the reference triangle and read each function of $\theta$ in terms of $x$ .
5 Go back to $x$ and conclude.

Concluding sentence

The triangle gives $\sin\theta=\frac{x}{3}$ and $\cos\theta=\frac{\sqrt{9-x^{2}}}{3}$ , so the integral equals $-\frac{\sqrt{9-x^{2}}}{9x}+C$ .

The trap: Not justifying that $\cos\theta\ge 0$ . The substitution $x=a\sin\theta$ is made on $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , where cosine is nonnegative: that is what allows you to write $\sqrt{a^{2}\cos^{2}\theta}=a\cos\theta$ without an absolute value.

Marking: Usually 1 mark for the announced substitution, 1 mark for the differential and the root, 2 marks for the computation in $\theta$ , 1 mark for going back to $x$ through the triangle.

Check before you hand in

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

Differentiate the answer

As in the previous chapter, differentiate the antiderivative you found: it must give back the integrand. It takes longer here, but it is always shorter than redoing the computation.

$xe^{x}-e^{x}$ differentiates into $e^{x}+xe^{x}-e^{x}=xe^{x}$ ✓. With the plus sign in the integration by parts formula, you would have obtained $xe^{x}+2e^{x}$ , which does not work.
Check that the new integral is simpler

After an integration by parts, compare $\int v\,du$ with $\int u\,dv$ . If the new one is more complicated, the choice of $u$ was reversed and you must start again.

$\int x\ln x\,dx$ with $u=x$ and $dv=\ln x\,dx$ first requires $v=x\ln x-x$ , itself an integration by parts, and the new integral $\int\left(x\ln x-x\right)dx$ is no simpler than the original one, which it contains: the signal is clear.
Test a partial fraction decomposition on one value

Substitute a simple number into the original fraction and into the decomposition: both must give the same result.

$\frac{1}{x(x-2)^{2}}$ at $x=1$ equals $1$ ; the decomposition $\frac{0.25}{x}-\frac{0.25}{x-2}+\frac{0.5}{(x-2)^{2}}$ equals $0.25+0.25+0.5=1$ there ✓.
Check the domain after a trigonometric substitution

Check that the interval chosen for $\theta$ makes the function under the square root nonnegative, which is what allows you to drop the absolute value.

With $x=3\sin\theta$ and $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ , $\cos\theta\ge 0$ , so $\sqrt{9\cos^{2}\theta}=3\cos\theta$ ✓.
Compare two forms that look different

Two correct answers can differ by a constant. Evaluate both at the same point and compare the DIFFERENCE: if it is constant, both are valid.

$\int\sin x\cos x\,dx$ equals $\frac{\sin^{2}x}{2}$ or $-\frac{\cos^{2}x}{2}$ , which differ by $\frac{1}{2}$ everywhere ✓.

The typical problem, taken apart

A complete trigonometric substitution

Compute $\displaystyle\int\frac{dx}{x^{2}\sqrt{9-x^{2}}}$ .

The triangle built on

x=3\sin\theta

: opposite side

x

, hypotenuse

3

, adjacent side

\sqrt{9-x^{2}}

. Every function of

\theta

can be read off it without any identity.

Step 1

The root has the form $\sqrt{a^{2}-x^{2}}$ with $a=3$ , and there is no lone $x$ in the numerator for a simple substitution. So set $x=3\sin\theta$ , with $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ .

Why

The interval for $\theta$ is not decoration: it is what guarantees $\cos\theta>0$ and allows you to write the root without an absolute value in the next step.

Step 2

$dx=3\cos\theta\,d\theta$ , $x^{2}=9\sin^{2}\theta$ , and $\sqrt{9-x^{2}}=\sqrt{9\left(1-\sin^{2}\theta\right)}=3\cos\theta$ .

Why

The identity $1-\sin^{2}\theta=\cos^{2}\theta$ is exactly what the substitution came for: it makes the root disappear, which no algebraic manipulation could do.

Step 3

The integral becomes $\int\frac{3\cos\theta\,d\theta}{9\sin^{2}\theta\cdot 3\cos\theta}=\frac{1}{9}\int\frac{d\theta}{\sin^{2}\theta}=\frac{1}{9}\int\csc^{2}\theta\,d\theta$ .

Why

The $\cos\theta$ factors cancel, which is the sign that the substitution was the right one. An integral where they do not cancel almost always points to a misidentified form.

Step 4

$\int\csc^{2}\theta\,d\theta=-\cot\theta+C$ , so the integral equals $-\frac{\cot\theta}{9}+C$ .

Why

This antiderivative is the only formula you need for this whole computation; it is the counterpart of $\int\sec^{2}\theta\,d\theta=\tan\theta$ .

Step 5

Reference triangle: $\sin\theta=\frac{x}{3}$ , so the opposite side is $x$ , the hypotenuse $3$ , the adjacent side $\sqrt{9-x^{2}}$ . Hence $\cot\theta=\frac{\sqrt{9-x^{2}}}{x}$ .

Why

No identity was needed: the triangle gives the cotangent directly. That is why the instruction is to DRAW it, not to deduce it.

Step 6

Answer: $-\frac{\sqrt{9-x^{2}}}{9x}+C$ . Numerical check at $x=1$ : the difference quotient of this antiderivative equals $0.3536$ , and $\frac{1}{1\cdot\sqrt{8}}=0.3536$ ✓.

Why

A numerical check at one point closes the computation in ten seconds and catches a sign error as well as a misplaced factor $9$ .

The conclusion, written out

$\displaystyle\int\frac{dx}{x^{2}\sqrt{9-x^{2}}}=-\frac{\sqrt{9-x^{2}}}{9x}+C$ , valid on each of the intervals $\left(-3, 0\right)$ and $\left(0, 3\right)$ .

The classic mistake on this problem: Replacing $\theta$ with $\arcsin\frac{x}{3}$ and leaving $-\frac{\cot\left(\arcsin\frac{x}{3}\right)}{9}$ . The expression is exact but unusable, and the triangle turns it into an algebraic fraction in three seconds.

Learn by heart

• Test SUBSTITUTION first: ten seconds, and it settles half of all integrals.
• $\int u\,dv=uv-\int v\,du$ , with a MINUS sign. $u$ is what simplifies when differentiated.
• LIATE: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.
• Partial fractions: divide first if the degree of the top is not strictly lower.
• A factor repeated to the power $n$ : $n$ terms. A quadratic factor: numerator $Ax+B$ .
• $\sqrt{a^{2}-x^{2}}\to x=a\sin\theta$ ; $\sqrt{a^{2}+x^{2}}\to x=a\tan\theta$ ; $\sqrt{x^{2}-a^{2}}\to x=a\sec\theta$ .
• Go back to $x$ through the reference TRIANGLE, never through an inverse-function identity.
• An $x$ in the numerator changes everything: $\frac{1}{x^{2}+1}$ gives an arctangent, $\frac{x}{x^{2}+1}$ a logarithm.

Frequently asked questions

How do you choose the right integration technique?

Test substitution first, by looking for an inner block whose derivative is present. If that fails, a product of different kinds of functions calls for integration by parts, a quotient of polynomials calls for partial fractions, and a root like the square root of a squared minus x squared calls for a trigonometric substitution.

How do you choose u in an integration by parts?

Follow the LIATE order: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. The first item of this list present in the integrand becomes u. The rule that sums it up is simple: u is what simplifies when differentiated, and dv is what you know how to integrate.

Why do you have to divide before decomposing into partial fractions?

Because partial fraction decomposition only applies to proper rational functions, whose numerator has a degree strictly lower than that of the denominator. Without dividing first, the system for the coefficients has no solution at all, and the computation gets stuck without any arithmetic mistake being the cause.

How many terms for a repeated factor in the denominator?

As many as the power. A squared factor gives two terms, one with the simple factor in the denominator and one with the squared factor. A cubed factor gives three. Forgetting a term makes the system impossible to solve, which is the most frequent dead end of the chapter.

How do you go back to the original variable after a trigonometric substitution?

By drawing the reference right triangle. If the substitution was x equals three sine theta, the opposite side is x and the hypotenuse is three, so the adjacent side is the square root of nine minus x squared. Every trigonometric function of theta can then be read directly off this triangle.

Practise it

Corrected exercises: Integration techniques, 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.

17 corrected exercises
170 points
225 minutes

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

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

The essentials

The three trigonometric substitutions, and their triangle

Integration by parts, and the choice of uuu

Partial fractions, in order

The mistakes that cost marks

1. Launching an integration by parts where a substitution is enough

2. Swapping the roles of uuu and dvdvdv

3. Losing the sign in the integration by parts formula

4. Decomposing into partial fractions without dividing first

5. Forgetting a term for a repeated factor

6. Confusing two integrands that differ only by an xxx

7. Going back to xxx without drawing the triangle

8. Not completing the square before substituting

Which method to choose

Which technique, according to the form of the integrand

Which uuu in an integration by parts

How the answer is expected to be written

A trigonometric substitution, written up for the marker

Check before you hand in

The typical problem, taken apart

A complete trigonometric substitution

Learn by heart

Frequently asked questions

How do you choose the right integration technique?

How do you choose u in an integration by parts?

Why do you have to divide before decomposing into partial fractions?

How many terms for a repeated factor in the denominator?

How do you go back to the original variable after a trigonometric substitution?

Corrected exercises: Integration techniques, 201-SN3-RE

See also

Looking for a 201-SN3-RE tutor in Montreal?

Integration by parts, and the choice of $u$

2. Swapping the roles of $u$ and $dv$

6. Confusing two integrands that differ only by an $x$

7. Going back to $x$ without drawing the triangle

Which $u$ in an integration by parts