MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: trigonometric substitution (MATH 141)

This sheet is not a summary of section 7.3 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on trigonometric substitution in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every number below is done by hand, as on the midterm and the final, and every antiderivative quoted has been checked the only way that settles it: by differentiating it back to the integrand.

The thread of the chapter

A trigonometric substitution is a round trip: the FORM under the root picks the identity, the interval of $\theta$ fixes the SIGN of the root once the square comes out, and the answer comes back to $x$ through the triangle, or stays in $\theta$ because the limits were changed. Every mark lost in this chapter is a missing leg of that trip.

This chapter is part of MATH 141, Calculus 2 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Trigonometric substitution
Pythagorean identities
Sine substitution
Tangent substitution
Secant substitution
Reference right triangle
Completing the square
Change of limits of integration
Half-angle formulas
Area of an ellipse
Area of a circular segment
Inverse trigonometric functions

The essentials

Three forms, three substitutions, one triangle

• $\sqrt{a^2 - x^2}$ : $x = a\sin\theta$ , $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ , $dx = a\cos\theta\,d\theta$ , and the root is $a\cos\theta$ because $\cos\theta \ge 0$ there.
• $\sqrt{a^2 + x^2}$ : $x = a\tan\theta$ , $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ , $dx = a\sec^2\theta\,d\theta$ , and the root is $a\sec\theta$ because $\sec\theta > 0$ there.
• $\sqrt{x^2 - a^2}$ : $x = a\sec\theta$ , $dx = a\sec\theta\tan\theta\,d\theta$ , and the root is $a|\tan\theta|$ : take $0 \le \theta < \frac{\pi}{2}$ for $x \ge a$ and $\pi \le \theta < \frac{3\pi}{2}$ for $x \le -a$ , where $\tan\theta \ge 0$ .
• Return to $x$ : draw the right triangle the substitution describes ( $\sin\theta = \frac{x}{a}$ puts $x$ opposite and $a$ on the hypotenuse). Pythagoras gives the third side, and it is always the root.
• Definite integral: move the limits with the substitution, $x = a \Rightarrow \theta = \frac{\pi}{2}$ for the sine, and never return to $x$ .

Where the root sits in each triangle: ADJACENT for the sine, HYPOTENUSE for the tangent, OPPOSITE for the secant. Every return to

x

reads two of these sides.

The FORM under the root chooses among the three substitutions and nothing else does. Whether a trigonometric substitution is needed at all is a separate question, asked first: an odd power of $x$ outside the root means $u =$ radicand is shorter.

Before substituting: the gestures that decide the form

• A quadratic with an $x$ term is completed first, minus sign factored out: $3 + 2x - x^2 = -(x^2 - 2x) + 3 = 4 - (x - 1)^2$ , then $x - 1 = 2\sin\theta$ .
• A coefficient in front of $x^2$ goes into the substitution: for $\sqrt{9 - 4x^2}$ , set $2x = 3\sin\theta$ , so $\sqrt{9 - 4x^2} = 3\cos\theta$ .
• Powers are handled like roots: $(x^2 + 4)^{3/2} = 8\sec^3\theta$ and $(x^2 + 4)^2 = 16\sec^4\theta$ with $x = 2\tan\theta$ .
• The $\theta$ integral usually needs section 7.2: $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ , $\tan^2\theta = \sec^2\theta - 1$ , $\int\sec^3\theta\,d\theta$ .
• Anything of the form $\sin 2\theta$ is expanded as $2\sin\theta\cos\theta$ BEFORE it is read on the triangle.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Back to x after x = 3 sin theta: what each expression becomes

Read a line as: after $x = 3\sin\theta$ , the $\theta$ expression of the first column is rewritten through the triangle (hypotenuse $3$ , opposite $x$ , adjacent $\sqrt{9 - x^2}$ ). Every example tests the line at $x = \frac{3}{2}$ , where $\theta = \frac{\pi}{6}$ . The red line is the rule students invent.

In theta	On the triangle	In x
$\theta$	$\sin\theta = \frac{x}{3}$	$\arcsin\frac{x}{3}$
Example: At $x = \frac{3}{2}$ : $\arcsin\frac{1}{2} = \frac{\pi}{6}$ .
$\cos\theta$	$\frac{\text{adj}}{\text{hyp}}$	$\frac{\sqrt{9 - x^2}}{3}$
Example: At $x = \frac{3}{2}$ : $\frac{1}{3}\sqrt{\frac{27}{4}} = \frac{\sqrt 3}{2} = \cos\frac{\pi}{6}$ .
$\tan\theta$	$\frac{\text{opp}}{\text{adj}}$	$\frac{x}{\sqrt{9 - x^2}}$
Example: At $x = \frac{3}{2}$ : $\frac{3/2}{3\sqrt 3/2} = \frac{1}{\sqrt 3} = \tan\frac{\pi}{6}$ .
$\sin 2\theta$	$2\sin\theta\cos\theta$	$\frac{2x\sqrt{9 - x^2}}{9}$
Example: At $x = \frac{3}{2}$ : $\frac{2\cdot\frac{3}{2}\cdot\frac{3\sqrt 3}{2}}{9} = \frac{\sqrt 3}{2} = \sin\frac{\pi}{3}$ .
$\cos 2\theta$	$1 - 2\sin^2\theta$	$1 - \frac{2x^2}{9}$
Example: At $x = \frac{3}{2}$ : $1 - \frac{2}{9}\cdot\frac{9}{4} = \frac{1}{2} = \cos\frac{\pi}{3}$ .
$\sin 2\theta$	$2\sin\theta$	$\frac{2x}{3}$ no such rule
Example: At $x = 3$ : $\theta = \frac{\pi}{2}$ and $\sin 2\theta = \sin\pi = 0$ , but $\frac{2x}{3} = 2$ , which no sine can equal. What to do: Expand $\sin 2\theta = 2\sin\theta\cos\theta$ first, then read both factors on the triangle.

Every blue line is a ratio of TWO sides of the triangle, or a double-angle formula expanded into such ratios. A single side, $\sqrt{9 - x^2}$ alone, is never a trigonometric value.

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. Choosing the substitution from the letters instead of the sign

the whole question

What not to write

“ $\sqrt{x^2 + 9}$ : I set $x = 3\sin\theta$ , so $\sqrt{9\sin^2\theta + 9}$ and then...”

What to write

“ $\sqrt{x^2 + 9}$ is a SUM, so $x = 3\tan\theta$ and $\sqrt{9\tan^2\theta + 9} = 3\sec\theta$ .”

Why: The substitution exists to trigger one identity. $9\sin^2\theta + 9$ matches none, so nothing collapses. Minus $x^2$ : sine. Plus: tangent. $x^2$ minus: secant.

2. Forgetting dx

2 to 3 marks, and every later step

What not to write

“ $x = 2\tan\theta$ , so $\int\frac{dx}{x^2 + 4} = \int\frac{d\theta}{4\sec^2\theta}$ .”

What to write

“ $dx = 2\sec^2\theta\,d\theta$ , so $\int\frac{2\sec^2\theta\,d\theta}{4\sec^2\theta} = \frac{\theta}{2} + C = \frac{1}{2}\arctan\frac{x}{2} + C$ .”

Why: A substitution replaces $x$ AND $dx$ . Here the forgotten $2\sec^2\theta$ is exactly what cancels the denominator; without it the integral becomes a harder one that has nothing to do with the question.

3. Keeping the old limits after the variable changed

the final answer, 2 to 3 marks

What not to write

“With $x = 2\sin\theta$ , $\int_0^1\frac{dx}{\sqrt{4 - x^2}} = \int_0^1 d\theta = 1$ .”

What to write

“ $x = 0 \Rightarrow \theta = 0$ and $x = 1 \Rightarrow \theta = \frac{\pi}{6}$ , so the integral is $\int_0^{\pi/6}d\theta = \frac{\pi}{6}$ .”

Why: Limits belong to the variable of integration. Write the new limits on the same line as the substitution, and never return to $x$ afterwards: the definite integral is finished in $\theta$ .

4. Taking the root of the secant positive on the negative branch

the whole question, since the answer has the wrong sign

What not to write

“ $\sqrt{x^2 - 4} = 2\tan\theta$ , so $\int_{-4}^{-2}\frac{\sqrt{x^2 - 4}}{x}\,dx = 2\sqrt 3 - \frac{2\pi}{3}$ .”

What to write

“With $\frac{\pi}{2} < \theta \le \pi$ , $\tan\theta \le 0$ , so $\sqrt{x^2 - 4} = -2\tan\theta$ and the integral is $\frac{2\pi}{3} - 2\sqrt 3$ , negative like its integrand.”

The integrand is odd: over

[-4, -2]

the curve is BELOW the axis, so the integral must be negative,

\frac{2\pi}{3} - 2\sqrt 3

. A positive answer is a sign error on the root.

Why: $\sqrt{4\tan^2\theta} = 2|\tan\theta|$ . On the principal range of arcsecant for $x \le -2$ the tangent is negative; either keep the absolute value or use $\pi \le \theta < \frac{3\pi}{2}$ , where it is positive.

5. Reading one side of the triangle as a trigonometric value

1 to 2 marks, the answer off by a factor 3

What not to write

“ $\sin\theta = \frac{x}{3}$ , so $\cos\theta = \sqrt{9 - x^2}$ .”

What to write

“ $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{9 - x^2}}{3}$ .”

Why: A trigonometric ratio is a ratio of TWO sides. The slip shows at once in a check: $\cos\theta \le 1$ , while $\sqrt{9 - x^2}$ reaches $3$ at $x = 0$ .

6. Leaving theta in an indefinite answer

the return mark, 1 to 2 marks

What not to write

“ $\int\sqrt{9 - x^2}\,dx = \frac{9}{2}\theta + \frac{9}{4}\sin 2\theta + C$ .”

What to write

“ $= \frac{9}{2}\arcsin\frac{x}{3} + \frac{x}{2}\sqrt{9 - x^2} + C$ , using $\sin 2\theta = 2\sin\theta\cos\theta = \frac{2x\sqrt{9 - x^2}}{9}$ .”

Why: An indefinite integral asked in $x$ is answered in $x$ . And $\sin\left(2\arcsin\frac{x}{3}\right)$ , left unsimplified, is not accepted either: expand the double angle and read the triangle.

7. Completing the square with the wrong sign

the whole question

What not to write

“ $3 + 2x - x^2 = (x - 1)^2 - 4$ , so I set $x - 1 = 2\sec\theta$ .”

What to write

“ $3 + 2x - x^2 = -\left[(x - 1)^2 - 1\right] + 3 = 4 - (x - 1)^2$ , so $x - 1 = 2\sin\theta$ .”

Why: Factor out the minus sign BEFORE completing. One value settles it: at $x = 1$ the left side is $4$ and the false right side is $-4$ . The wrong sign also moves you onto a domain where the root does not exist.

8. Dropping the absolute value without a reason

1 mark of justification

What not to write

“ $\sqrt{9 - 9\sin^2\theta} = 3\cos\theta$ .”

What to write

“ $\sqrt{9 - 9\sin^2\theta} = 3|\cos\theta| = 3\cos\theta$ , since $\cos\theta \ge 0$ for $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ .”

Why: $\sqrt{u^2} = |u|$ , always. The interval of $\theta$ is chosen precisely so that the absolute value can be dropped, and saying so is what the marker is looking for. The same sentence, with the tangent, is what saves the secant on its negative branch.

Which method to choose

Which substitution, by the FORM of the integrand

Look at what stands outside the root, then at the sign pattern under it, before writing any substitution

With

a = 2

\sqrt{4 - x^2}

lives on

[-2, 2]

(sine),

\sqrt{x^2 + 4}

on every real (tangent),

\sqrt{x^2 - 4}

on two separate pieces (secant, one interval of

\theta

per piece).

If an odd power of $x$ multiplies the root → plain substitution $u =$ the radicand

Example: $\int x\sqrt{16 - x^2}\,dx = -\frac{1}{3}(16 - x^2)^{3/2} + C$
If $\frac{1}{\sqrt{a^2 - x^2}}$ or $\frac{1}{a^2 + x^2}$ alone → no substitution: the table of antiderivatives

Example: $\int_0^1\frac{dx}{\sqrt{4 - x^2}} = \arcsin\frac{1}{2} = \frac{\pi}{6}$
If $a^2 - x^2$ : constant first, minus $x^2$ → $x = a\sin\theta$ , and $x$ stays in $[-a, a]$

Example: $\int_0^3 x^2\sqrt{9 - x^2}\,dx = \frac{81\pi}{16}$
If $a^2 + x^2$ , under a root or raised to any power → $x = a\tan\theta$ , every real $x$

Example: $\int_0^2\frac{dx}{(x^2 + 4)^{3/2}} = \frac{1}{4}\sin\frac{\pi}{4} = \frac{\sqrt 2}{8}$
If $x^2 - a^2$ : $x^2$ first, minus a constant → $x = a\sec\theta$ , and check the sign of $\tan\theta$ on the branch

Example: $\int_1^2\frac{\sqrt{x^2 - 1}}{x}\,dx = \sqrt 3 - \frac{\pi}{3}$
If a quadratic with an $x$ term → complete the square, then one of the three forms on $u = x - h$

Example: $x^2 + 6x + 13 = (x + 3)^2 + 4$ : $x + 3 = 2\tan\theta$

The first two branches are checked BEFORE the three forms: a trigonometric substitution that a plain substitution or the table could avoid costs five minutes and adds a triangle to get wrong. Partial fractions are the next chapter; nothing here needs them.

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 a trigonometric substitution the marker can follow

When to use it: Any integral with $a^2 - x^2$ , $a^2 + x^2$ or $x^2 - a^2$ that neither a plain substitution nor the table handles

1 Name the form and the substitution on one line, with its interval: “ $\sqrt{9 - x^2}$ , so $x = 3\sin\theta$ , $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ .”
2 Write $dx$ and the root, the sign justified by the interval: $dx = 3\cos\theta\,d\theta$ and $\sqrt{9 - x^2} = 3|\cos\theta| = 3\cos\theta$ .
3 Integrate in $\theta$ , naming the tool of section 7.2 used: half-angle formula, $\tan^2\theta = \sec^2\theta - 1$ , $\int\sec^3\theta\,d\theta$ .
4 Indefinite: return through the triangle, drawn with its three sides labelled. Definite: move the limits on the line of the substitution and finish in $\theta$ .
5 Check: differentiate an indefinite answer; compare the sign and size of a definite one with the integrand.

Concluding sentence

“With $x = 3\sin\theta$ , $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ , we have $\sqrt{9 - x^2} = 3\cos\theta$ since $\cos\theta \ge 0$ on this interval, so $\int\sqrt{9 - x^2}\,dx = \frac{9}{2}\arcsin\frac{x}{3} + \frac{x}{2}\sqrt{9 - x^2} + C$ .”

The trap: Skipping the triangle and writing $\cos\theta$ from memory: the division by the hypotenuse is what goes missing.

Marking: Typically 1 mark for the substitution with dx, 1 for the root simplified with its sign, 2 for the integral in theta, 1 for the return to x or the new limits.

Check before you hand in

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

Differentiate the answer

One minute, and the only proof that the triangle was read correctly. Differentiating is always easier than integrating.

$\frac{d}{dx}\left[-\frac{\sqrt{25 - x^2}}{25x}\right] = \frac{1}{x^2\sqrt{25 - x^2}}$ : consistent. A leftover factor of $3$ or $\frac{1}{3}$ means a side was read without the hypotenuse.
The sign of a definite integral

Look at the sign of the integrand on the interval before computing: the answer must have the same sign.

$\int_{-2}^{-1}\frac{\sqrt{x^2 - 1}}{x}\,dx$ must be negative; $\frac{\pi}{3} - \sqrt 3 < 0$ is consistent, $\sqrt 3 - \frac{\pi}{3} > 0$ is a sign error on the root.
Test one value on the triangle

Pick an $x$ with a known angle and compare the $\theta$ expression with its $x$ version.

At $x = \frac{3}{2}$ , $\theta = \frac{\pi}{6}$ : $\frac{2x\sqrt{9 - x^2}}{9} = \frac{\sqrt 3}{2} = \sin\frac{\pi}{3}$ . The false $\frac{2x}{3} = 1$ fails.
Bracket with rectangles

The integrand between its smallest and largest value, times the length of the interval, gives two bounds with no calculator.

$\int_0^2\frac{dx}{(x^2 + 4)^{3/2}} \le 2\cdot\frac{1}{8} = \frac{1}{4}$ , and $\frac{\sqrt 2}{8} \approx 0.18$ : consistent.
Known areas at the ends

An antiderivative of $\sqrt{a^2 - x^2}$ must give $\frac{\pi a^2}{2}$ from $-a$ to $a$ : the half disc.

$\left[\frac{9}{2}\arcsin\frac{x}{3} + \frac{x}{2}\sqrt{9 - x^2}\right]_{-3}^{3} = \frac{9\pi}{2}$ : consistent.

The typical problem, taken apart

Completing the square, then the secant: an integral with a power 3/2

Compute $\displaystyle\int\frac{dx}{(x^2 + 2x)^{3/2}}$ for $x > 0$ .

No calculator. Every step must be justified as on a MATH 141 final.

Step 1

$x^2 + 2x = (x + 1)^2 - 1$ . For $x > 0$ , $u = x + 1 > 1$ , and the form is $u^2 - 1$ : a secant, $x + 1 = \sec\theta$ with $0 < \theta < \frac{\pi}{2}$ .

Why

The $x$ term rules out a direct substitution; completing the square reveals the form. Stating $x > 0$ places $u$ on the positive branch, which is what makes $\tan\theta > 0$ legitimate.

Step 2

$dx = \sec\theta\tan\theta\,d\theta$ and $(x^2 + 2x)^{3/2} = (\sec^2\theta - 1)^{3/2} = (\tan^2\theta)^{3/2} = \tan^3\theta$ , since $\tan\theta > 0$ on $(0, \frac{\pi}{2})$ .

Why

The power $\frac{3}{2}$ is a root cubed: $(\tan^2\theta)^{3/2} = |\tan\theta|^3$ , and the interval is what drops the absolute value. That sentence is worth a mark.

Step 3

$\int\frac{\sec\theta\tan\theta}{\tan^3\theta}\,d\theta = \int\frac{\sec\theta}{\tan^2\theta}\,d\theta = \int\frac{\cos\theta}{\sin^2\theta}\,d\theta = -\frac{1}{\sin\theta} + C$ , with $w = \sin\theta$ .

Why

A quotient of $\sec$ and $\tan$ that matches no formula is rewritten in sines and cosines; then $w = \sin\theta$ turns it into $\int w^{-2}\,dw$ .

Step 4

The triangle of $\sec\theta = \frac{x + 1}{1}$ : hypotenuse $x + 1$ , adjacent $1$ , opposite $\sqrt{(x + 1)^2 - 1} = \sqrt{x^2 + 2x}$ . So $\frac{1}{\sin\theta} = \frac{x + 1}{\sqrt{x^2 + 2x}}$ .

Why

The sides are labelled with $x + 1$ , never with $x$ : the shift of the completed square survives to the last line. The opposite side is, as always for the secant, the root.

Step 5

Check: $\frac{d}{dx}\left[-\frac{x + 1}{\sqrt{x^2 + 2x}}\right] = \frac{-(x^2 + 2x) + (x + 1)^2}{(x^2 + 2x)^{3/2}} = \frac{1}{(x^2 + 2x)^{3/2}}$ .

Why

The quotient rule gives $-\frac{\sqrt{x^2 + 2x} - (x + 1)\frac{x + 1}{\sqrt{x^2 + 2x}}}{x^2 + 2x}$ , and $(x + 1)^2 - (x^2 + 2x) = 1$ closes it. One minute, and every step above is confirmed.

The conclusion, written out

“For $x > 0$ , $\int\frac{dx}{(x^2 + 2x)^{3/2}} = -\frac{x + 1}{\sqrt{x^2 + 2x}} + C$ .”

The classic mistake on this problem: Setting $x = \sec\theta$ on $x^2 + 2x$ before completing the square, which collapses nothing; or labelling the triangle with $x$ on the hypotenuse and returning $-\frac{x}{\sqrt{x^2 - 1}}$ , the answer to a different integral.

Learn by heart

• $a^2 - x^2$ : $x = a\sin\theta$ . $a^2 + x^2$ : $x = a\tan\theta$ . $x^2 - a^2$ : $x = a\sec\theta$ .
• $\sqrt{u^2} = |u|$ : the interval of $\theta$ decides the sign, and you SAY so.
• The root is the adjacent side for the sine, the hypotenuse for the tangent, the opposite side for the secant.
• Definite integral: move the limits with the substitution, never return to $x$ .
• $\sin 2\theta = 2\sin\theta\cos\theta$ BEFORE reading the triangle.
• Odd power of $x$ outside the root: $u =$ radicand first. Quadratic with an $x$ term: complete the square first.
• $\int\sqrt{a^2 - x^2}\,dx = \frac{a^2}{2}\arcsin\frac{x}{a} + \frac{x}{2}\sqrt{a^2 - x^2} + C$ . Ellipse: $\pi ab$ .

Frequently asked questions

How do I know which trigonometric substitution to use?

Look at the sign pattern under the root. A constant minus x squared calls for x equals a sine theta, a constant plus x squared for x equals a tangent theta, and x squared minus a constant for x equals a secant theta. Each choice triggers one Pythagorean identity that removes the root. If the quadratic has an x term, complete the square first.

Do I have to change the limits of integration in a trig substitution?

Yes, if you finish the computation in theta, which is the shortest route. Convert each limit with the substitution itself: with x equals 2 sine theta, x equals 1 becomes theta equals pi over 6. Then never return to x. Keeping the old limits on the theta integral is one of the most common ways to lose the final answer.

How do I get back to x after a trigonometric substitution?

Draw the right triangle the substitution describes. For x equals 3 sine theta, put x on the side opposite theta and 3 on the hypotenuse; Pythagoras gives the adjacent side, the square root of 9 minus x squared. Read every ratio as a ratio of two sides, and expand a double angle like sine 2 theta into 2 sine theta cosine theta before reading it.

When should I not use a trig substitution?

When an odd power of x sits outside the root, as in x times the root of 16 minus x squared: the plain substitution u equals the radicand is done in two lines. Also when the integral is already in the table, like 1 over the root of a squared minus x squared, which is arcsine of x over a. Check these two cases before drawing any triangle.

Why does the sign of the square root matter with x = a sec theta?

Because the root of a squared tangent squared is a times the absolute value of tangent theta. For x at least a, theta is in the first quadrant and the tangent is positive. For x at most minus a, with the principal range of arcsecant, the tangent is negative, so the root is minus a tangent theta. Forgetting it gives an answer with the wrong sign.

Practise it

Corrected exercises: Trigonometric substitution, MATH 141 at McGill

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.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Trigonometric integrals Next sheet Partial fractions

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Revision sheet: trigonometric substitution (MATH 141)

The essentials

Three forms, three substitutions, one triangle

Before substituting: the gestures that decide the form

The rules in table form

Back to x after x = 3 sin theta: what each expression becomes

The mistakes that cost marks

1. Choosing the substitution from the letters instead of the sign

2. Forgetting dx

3. Keeping the old limits after the variable changed

4. Taking the root of the secant positive on the negative branch

5. Reading one side of the triangle as a trigonometric value

6. Leaving theta in an indefinite answer

7. Completing the square with the wrong sign

8. Dropping the absolute value without a reason

Which method to choose

Which substitution, by the FORM of the integrand

How the answer is expected to be written

Writing a trigonometric substitution the marker can follow

Check before you hand in

The typical problem, taken apart

Completing the square, then the secant: an integral with a power 3/2

Learn by heart

Frequently asked questions

How do I know which trigonometric substitution to use?

Do I have to change the limits of integration in a trig substitution?

How do I get back to x after a trigonometric substitution?

When should I not use a trig substitution?

Why does the sign of the square root matter with x = a sec theta?

Corrected exercises: Trigonometric substitution, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?