MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: trigonometric integrals (MATH 141)

This sheet is not a summary of section 7.2 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on trigonometric integrals in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every antiderivative below has been checked by differentiating it back, and every value is exact, as on the calculator-free midterm and final. The method is always the same three lines: which factor is saved, which identity converts the rest, which substitution follows.

The thread of the chapter

A trigonometric integral is a substitution in disguise: save ONE factor to be the $du$ , take it from the ODD power, and convert everything else with a Pythagorean identity. When every power is even there is nothing to save, and the half-angle formulas lower the degree instead.

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 integral
Powers of sine and cosine
Powers of tangent and secant
Pythagorean identities
Half-angle formulas
Integral of the secant
Integral of secant cubed
Product-to-sum formulas
Orthogonality of sines and cosines
Substitution rule for definite integrals

The essentials

Save one factor from the odd power, convert the rest

• The four possible differentials: $\cos x\,dx = d(\sin x)$ , $\sin x\,dx = -d(\cos x)$ , $\sec^2 x\,dx = d(\tan x)$ , $\sec x\tan x\,dx = d(\sec x)$ .
• Detach ONE of them, then convert every remaining factor into the new variable with $\sin^2 x + \cos^2 x = 1$ or $\tan^2 x = \sec^2 x - 1$ . What is left must be a polynomial (or a power) in $u$ .
• The detached factor comes from the ODD power, because the power left behind is then even and converts through squares. An odd leftover converts into a square root.
• The power of the OTHER function does not matter: even, odd, negative or fractional, it is carried along as $u^k$ ( $\sqrt{\cos x}$ becomes $u^{1/2}$ ).
• When both powers are odd, both routes work; the answers differ by a constant. Save from the smaller power, the polynomial is shorter.

Write the three choices on the first line of the answer: the factor saved, the identity used, the substitution. That line alone is usually the method mark.

No factor to save: lower the degree

• If every power of $\sin x$ and $\cos x$ is EVEN, use $\sin^2 x = \frac{1 - \cos 2x}{2}$ , $\cos^2 x = \frac{1 + \cos 2x}{2}$ and $\sin x\cos x = \frac{\sin 2x}{2}$ .
• Each use halves the power and doubles the frequency. $\cos^4 x$ needs two uses, and the second one is applied to $\cos^2 2x$ .
• Halving the degree can create an ODD power of $\cos 2x$ : switch back to saving a factor for that term.
• Products of DIFFERENT frequencies, $\sin mx\cos nx$ , are neither powers nor substitutions: turn them into sums with the product-to-sum formulas first.
• Over whole periods, $\sin^2$ and $\cos^2$ average $\frac{1}{2}$ , and every mixed product of integer frequencies integrates to $0$ on $[-\pi, \pi]$ .

\sin^2 x

has HALF the period of

\sin x

and oscillates about

\frac{1}{2}

: that picture is the half-angle formula, a constant

\frac{1}{2}

plus a wave of frequency

2

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.

Which factor to save, read off the powers

Read a line as: when the integrand has this form, save this factor, and the substitution in the last column finishes the job. The red lines are the forms where NO factor can be saved.

Form of the integrand	Factor saved and identity	Substitution
$\sin^m x\cos^n x$ , $m$ odd	$\sin x\,dx$ ; $\sin^2 x = 1 - \cos^2 x$	$u = \cos x$
Example: $\int \sin^3 x\cos^4 x\,dx = -\int (1 - u^2)u^4\,du = -\frac{\cos^5 x}{5} + \frac{\cos^7 x}{7} + C$
$\sin^m x\cos^n x$ , $n$ odd	$\cos x\,dx$ ; $\cos^2 x = 1 - \sin^2 x$	$u = \sin x$
Example: $\int \cos^3 x\,dx = \int (1 - u^2)\,du = \sin x - \frac{\sin^3 x}{3} + C$
$\tan^m x\sec^n x$ , $n$ even	$\sec^2 x\,dx$ ; $\sec^2 x = 1 + \tan^2 x$	$u = \tan x$
Example: $\int \tan^2 x\sec^4 x\,dx = \int u^2(1 + u^2)\,du = \frac{\tan^3 x}{3} + \frac{\tan^5 x}{5} + C$
$\tan^m x\sec^n x$ , $m$ odd	$\sec x\tan x\,dx$ ; $\tan^2 x = \sec^2 x - 1$	$u = \sec x$
Example: $\int \tan^3 x\sec x\,dx = \int (u^2 - 1)\,du = \frac{\sec^3 x}{3} - \sec x + C$
$\sin^m x\cos^n x$ , $m$ and $n$ even	none	no substitution no factor to save
Example: $\int \sin^2 x\cos^2 x\,dx = \int \frac{\sin^2 2x}{4}\,dx = \frac{x}{8} - \frac{\sin 4x}{32} + C$ What to do: Lower the degree with the half-angle formulas, as many times as needed.
$\tan^m x\sec^n x$ , $m$ even, $n$ odd	none	no substitution no factor to save
Example: $\int \tan^2 x\sec x\,dx = \int \sec^3 x\,dx - \int \sec x\,dx = \frac{1}{2}(\sec x\tan x - \ln\|\sec x + \tan x\|) + C$ What to do: Rewrite everything in powers of $\sec x$ , then use $\int \sec x$ and $\int \sec^3 x$ (by parts).

When a line is blue for two reasons at once (odd power of $\tan x$ AND even power of $\sec x$ ), both substitutions work and give answers that differ by a constant.

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. Saving the factor from the even power

the whole question: the square root blocks the computation

What not to write

“ $\int \sin^3 x\cos^2 x\,dx$ : $u = \sin x$ , $du = \cos x\,dx$ , so it is $\int u^3\cos x\,du$ with $\cos x = \sqrt{1 - u^2}$ .”

What to write

“The odd power is on $\sin x$ : save $\sin x\,dx = -du$ with $u = \cos x$ , so $-\int (1 - u^2)u^2\,du = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C$ .”

Why: After detaching one factor, the power left of that function must be EVEN so that it converts through $\sin^2 = 1 - \cos^2$ . Detaching from $\cos^2 x$ leaves $\cos^1 x$ , a square root in $u$ .

2. The wrong sign in the half-angle formula

1 to 2 marks, and every later line

What not to write

“ $\cos^2 x = \frac{1 - \cos 2x}{2}$ , so $\int_0^{\pi/4}\cos^2 x\,dx = \frac{\pi}{8} - \frac{1}{4}$ .”

What to write

“ $\cos^2 x = \frac{1 + \cos 2x}{2}$ , so $\int_0^{\pi/4}\cos^2 x\,dx = \left[\frac{x}{2} + \frac{\sin 2x}{4}\right]_0^{\pi/4} = \frac{\pi}{8} + \frac{1}{4}$ .”

Why: Test the formula at $x = 0$ before using it: $\cos^2 0 = 1$ , and only $\frac{1 + \cos 0}{2}$ gives $1$ . The minus sign belongs to $\sin^2 x$ , which vanishes at $0$ .

3. Integrating a square with the power rule

the whole question

What not to write

“ $\int \sin^2 x\,dx = \frac{\sin^3 x}{3} + C$ .”

What to write

“All powers are even: $\int \sin^2 x\,dx = \int \frac{1 - \cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$ .”

Why: $\frac{d}{dx}\frac{\sin^3 x}{3} = \sin^2 x\cos x$ : the power rule needs $du = \cos x\,dx$ in the integrand. The false answer also gives $0$ on $[0, \pi]$ for a positive integrand whose integral is $\frac{\pi}{2}$ .

4. Replacing the square root by cos x where the cosine is negative

the whole question, with an impossible answer

What not to write

“ $u = \sin x$ , $\cos x = \sqrt{1 - u^2}$ , so $\int_0^\pi \cos^2 x\,dx = \int_0^0 \sqrt{1 - u^2}\,du = 0$ .”

What to write

“All powers are even, so no factor is saved: $\int_0^\pi \frac{1 + \cos 2x}{2}\,dx = \frac{\pi}{2}$ .”

[0, \frac{\pi}{2}]

the two curves coincide; beyond

\frac{\pi}{2}

\cos x

goes negative while

\sqrt{1 - \sin^2 x} = |\cos x|

stays positive. The substitution flips the sign of half the integral.

Why: $\sqrt{1 - \sin^2 x} = |\cos x|$ , which differs from $\cos x$ on $(\frac{\pi}{2}, \pi)$ . A positive integrand cannot have integral $0$ : that check alone catches it. The figure shows the two curves parting at $\frac{\pi}{2}$ .

5. Keeping the old lower bound after the substitution u = sec x

1 mark

What not to write

“ $\int_0^{\pi/3}\tan x\sec^3 x\,dx = \int_0^2 u^2\,du = \frac{8}{3}$ .”

What to write

“ $u = \sec x$ runs from $\sec 0 = 1$ to $\sec\frac{\pi}{3} = 2$ , so the integral is $\int_1^2 u^2\,du = \frac{7}{3}$ .”

Why: Unlike $\sin 0$ and $\tan 0$ , $\sec 0 = 1$ . Every new bound is computed, never copied.

6. Losing the one half in the integral of sec cubed

1 to 2 marks

What not to write

“ $I = \sec x\tan x - I + \ln|\sec x + \tan x|$ , so $\int \sec^3 x\,dx = \sec x\tan x + \ln|\sec x + \tan x| + C$ .”

What to write

“ $2I = \sec x\tan x + \ln|\sec x + \tan x|$ , so $\int \sec^3 x\,dx = \frac{1}{2}\left(\sec x\tan x + \ln|\sec x + \tan x|\right) + C$ .”

Why: The integral reappears on the right with a minus sign: move it to the left, then DIVIDE by $2$ , and add the constant only after. Differentiating the answer is the ten-second check.

7. Dropping the sign of sin(A - B) when A is smaller than B

1 mark, and the derivative check fails

What not to write

“ $\sin 2x\cos 5x = \frac{1}{2}[\sin 3x + \sin 7x]$ , so $\int \sin 2x\cos 5x\,dx = -\frac{\cos 3x}{6} - \frac{\cos 7x}{14} + C$ .”

What to write

“ $\sin 2x\cos 5x = \frac{1}{2}[\sin(-3x) + \sin 7x] = \frac{1}{2}[-\sin 3x + \sin 7x]$ , so the integral is $\frac{\cos 3x}{6} - \frac{\cos 7x}{14} + C$ .”

Why: Sine is odd, $\sin(-3x) = -\sin 3x$ ; cosine is even, so the same slip is harmless in $\sin A\sin B$ and $\cos A\cos B$ . The false answer is the antiderivative of $\sin 5x\cos 2x$ .

8. Copying the csc formula from the sec formula

1 mark

What not to write

“By symmetry, $\int \csc x\,dx = \ln|\csc x + \cot x| + C$ .”

What to write

“ $\int \csc x\,dx = -\ln|\csc x + \cot x| + C = \ln|\csc x - \cot x| + C$ .”

Why: The co-functions differentiate with a minus sign: $(\csc x)' = -\csc x\cot x$ and $(\cot x)' = -\csc^2 x$ . Differentiating the false answer gives $-\csc x$ .

Which method to choose

Which gesture, by the FORM of the integrand

Look at the functions and at the parity of each power before writing anything

If an inner function ( $\sqrt x$ , $x^2$ , $3x$ ) with its derivative outside → substitute for it first, then come back to this tree

Example: $\int \frac{\cos^3\sqrt x}{\sqrt x}\,dx = 2\int \cos^3 w\,dw$
If a quotient, or $\cot x$ , $\csc x$ mixed with other functions → rewrite everything in $\sin x$ and $\cos x$

Example: $\cos^2 x\tan^3 x = \frac{\sin^3 x}{\cos x}$ , odd power of $\sin x$
If powers of $\sin x$ and $\cos x$ , one of them odd → save one factor of the odd power, convert the rest

Example: $\sin^2 x\cos^5 x$ : $u = \sin x$
If powers of $\sin x$ and $\cos x$ , all even → lower the degree with the half-angle formulas

Example: $\cos^4 x = \frac{3}{8} + \frac{\cos 2x}{2} + \frac{\cos 4x}{8}$
If powers of $\tan x$ and $\sec x$ → $\sec$ even: save $\sec^2 x$ ; $\tan$ odd: save $\sec x\tan x$ ; otherwise rewrite in $\sec x$ alone

Example: $\tan^2 x\sec x = \sec^3 x - \sec x$
If $\sin mx\cos nx$ , $\sin mx\sin nx$ or $\cos mx\cos nx$ with $m \neq n$ → product to sum, then integrate each term

Example: $\sin 5x\cos 3x = \frac{1}{2}(\sin 2x + \sin 8x)$

On a definite integral, look for a symmetry before computing: an odd function on $[-a, a]$ gives $0$ , and $\int_0^{\pi/2}\sin^2 x\,dx = \int_0^{\pi/2}\cos^2 x\,dx = \frac{\pi}{4}$ without any formula.

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 integral for full marks

When to use it: Any integral of powers of $\sin$ , $\cos$ , $\tan$ or $\sec$ , indefinite or definite

1 Name the parity that decides: “the power of $\cos x$ is odd” or “all powers are even”.
2 Write the factor saved and the substitution: “save $\cos x\,dx$ , $u = \sin x$ , $du = \cos x\,dx$ ”.
3 Convert the rest with the identity written in full, $\cos^2 x = 1 - u^2$ , and write the integral entirely in $u$ before integrating.
4 For a definite integral, compute the new bounds ( $\sec 0 = 1$ , $\sin\pi = 0$ ) and never come back to $x$ .
5 Return to $x$ for an indefinite integral, add $+ C$ , and check by differentiating one term if time allows.

Concluding sentence

“The power of $\sin x$ is odd, so I save $\sin x\,dx = -du$ with $u = \cos x$ and write $\sin^2 x = 1 - u^2$ : the integral becomes $-\int (1 - u^2)u^4\,du$ .”

The trap: Integrating in $u$ while some $x$ is still in the integrand, or keeping the old bounds after substituting. Both are caught by writing the full integral in $u$ on one line before integrating.

Marking: Typically 1 mark for the correct choice of factor and substitution, 1 for the conversion, 1 for the integration, and 1 for the bounds or the return to x.

Check before you hand in

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

Differentiate one term back

The derivative of the answer must be the integrand. On a long polynomial in sin or cos, differentiating the leading term already catches a wrong factor or a wrong sign.

$\frac{d}{dx}\left(-\frac{2}{5}\cos^5 x\right) = 2\sin x\cos^4 x = \sin 2x\cos^3 x$ : consistent.
The sign of the integrand

A positive integrand on $[a, b]$ with $a < b$ has a positive integral. A result of $0$ or a negative number means a bound in the wrong order or a square root that flipped a sign.

$\int_0^\pi \sin^3 x\,dx = \frac{4}{3} > 0$ ; $\int_0^\pi \cos^2 x\,dx$ cannot be $0$ .
Test a formula at x = 0

Half-angle and product-to-sum formulas are checked in five seconds at a convenient point before being used.

$\cos^2 0 = 1 = \frac{1 + \cos 0}{2}$ ; $\sin\frac{\pi}{2}\sin\frac{3\pi}{2} = -1 = \frac{1}{2}[\cos\pi - \cos 2\pi]$ .
Two routes, one constant

If a classmate's answer looks different, subtract: two correct antiderivatives differ by a CONSTANT, which you can find by evaluating both at one point.

$\frac{\sec^4 3x}{12} - \left(\frac{\tan^2 3x}{6} + \frac{\tan^4 3x}{12}\right) = \frac{1}{12}$ for every $x$ .

The typical problem, taken apart

One integral, two legitimate substitutions

Evaluate $\int_0^{\pi/4}\tan^3 x\sec^4 x\,dx$ exactly, then evaluate it a second way and compare.

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

The region under

y = \tan^3 x\sec^4 x

[0, \frac{\pi}{4}]

; the curve climbs from

0

\tan^3\frac{\pi}{4}\sec^4\frac{\pi}{4} = 4

Step 1

The power of $\sec x$ is $4$ , even, AND the power of $\tan x$ is $3$ , odd: both rules apply. Route 1: save $\sec^2 x\,dx = du$ with $u = \tan x$ , and convert $\sec^2 x = 1 + u^2$ .

Why

Naming the parity is the first mark. When both rules apply, either route is correct; choose the one with the shorter polynomial if you can see it.

Step 2

Bounds: $\tan 0 = 0$ , $\tan\frac{\pi}{4} = 1$ . So $\int_0^{\pi/4}\tan^3 x\sec^2 x\,(\sec^2 x\,dx) = \int_0^1 u^3(1 + u^2)\,du = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}$ .

Why

The bounds are changed at the moment of the substitution, and $x$ never comes back. Two powers of $u$ , two fractions: nothing to expand.

Step 3

Route 2: save $\sec x\tan x\,dx = dv$ with $v = \sec x$ . What is left is $\tan^2 x\sec^3 x = (v^2 - 1)v^3$ .

Why

Count the factors: one $\sec x$ and one $\tan x$ are spent on $dv$ , so $\tan^2 x$ and $\sec^3 x$ remain, not $\tan^3$ and $\sec^4$ .

Step 4

Bounds: $\sec 0 = 1$ , $\sec\frac{\pi}{4} = \sqrt 2$ . So the integral is $\int_1^{\sqrt 2}(v^5 - v^3)\,dv = \left[\frac{v^6}{6} - \frac{v^4}{4}\right]_1^{\sqrt 2} = \left(\frac{8}{6} - 1\right) - \left(\frac{1}{6} - \frac{1}{4}\right) = \frac{1}{3} + \frac{1}{12} = \frac{5}{12}$ .

Why

The lower bound is $1$ , not $0$ . $(\sqrt 2)^6 = 8$ and $(\sqrt 2)^4 = 4$ : exact powers, no decimals.

Step 5

Check: the two routes agree, and on the figure the region sits under a curve that stays below $4$ on an interval of length $\frac{\pi}{4} \approx 0.79$ , so the area is below $3.2$ ; being mostly near $0$ on the left, $\frac{5}{12} \approx 0.42$ is plausible.

Why

Two independent routes giving the same number is the strongest check available without a calculator. The indefinite answers, $\frac{\tan^4 x}{4} + \frac{\tan^6 x}{6}$ and $\frac{\sec^6 x}{6} - \frac{\sec^4 x}{4}$ , differ by the constant $\frac{1}{12}$ .

The conclusion, written out

“ $\int_0^{\pi/4}\tan^3 x\sec^4 x\,dx = \frac{5}{12}$ , obtained with $u = \tan x$ and confirmed with $v = \sec x$ .”

The classic mistake on this problem: Keeping the lower bound $0$ in route 2, which gives $\frac{8}{6} - 1 = \frac{1}{3}$ ; or saving $\sec x\tan x$ and then converting $\tan^3 x$ instead of $\tan^2 x$ , which produces a square root.

Learn by heart

• Save ONE factor from the ODD power; convert the rest with $\sin^2 + \cos^2 = 1$ or $\tan^2 = \sec^2 - 1$ .
• All powers even: $\sin^2 x = \frac{1 - \cos 2x}{2}$ , $\cos^2 x = \frac{1 + \cos 2x}{2}$ , $\sin x\cos x = \frac{\sin 2x}{2}$ .
• $\sec$ even: save $\sec^2 x$ , $u = \tan x$ . $\tan$ odd: save $\sec x\tan x$ , $u = \sec x$ .
• $\int \sec x\,dx = \ln|\sec x + \tan x| + C$ ; $\int \sec^3 x\,dx = \frac{1}{2}(\sec x\tan x + \ln|\sec x + \tan x|) + C$ .
• $\int \tan x\,dx = \ln|\sec x| + C$ ; $\int \csc x\,dx = -\ln|\csc x + \cot x| + C$ .
• Different frequencies: product to sum first. $\sin(-\theta) = -\sin\theta$ , $\cos(-\theta) = \cos\theta$ .
• On $[-\pi, \pi]$ : mixed products of integer frequencies give $0$ ; $\sin^2 mx$ and $\cos^2 mx$ give $\pi$ .
• $\sqrt{1 - \sin^2 x} = |\cos x|$ , never $\cos x$ without checking its sign.

Frequently asked questions

How do I know which substitution to use for a power of sine times a power of cosine?

Look for the odd power. Detach one factor of it to be the differential, and convert all the other factors of that same function with sine squared plus cosine squared equals one. If the power of sine is odd, let u be cosine; if the power of cosine is odd, let u be sine. If both powers are even, no substitution works, and you lower the degree with the half-angle formulas instead.

When do I use the half-angle formulas in an integral?

Only when every power of sine and cosine is even, because then there is no factor you can save for a substitution. Replace sine squared by one minus cosine of 2x, all over 2, and cosine squared by one plus cosine of 2x, all over 2. Repeat until every term is a single cosine or a constant. A fourth power needs the formula twice.

What is the integral of secant cubed and how do I remember it?

Do not memorize it: rebuild it by parts in four lines. Take u equal to secant and dv equal to secant squared dx, replace tangent squared by secant squared minus one, and the original integral reappears with a minus sign. Move it to the left and divide by two. The answer is one half of secant times tangent plus the log of the absolute value of secant plus tangent, plus a constant.

Why do two correct methods give different answers for the same trig integral?

Because two antiderivatives of the same function on an interval can differ by a constant, and trigonometric identities hide that constant. The integral of sine cubed times cosine cubed gives one expression in sine and another in cosine, and they differ by exactly one twelfth. Subtract the two answers or evaluate both at one point: if the difference is a constant, both are correct.

Can I use a calculator for trigonometric integrals in MATH 141?

No. The midterm and the final are written without a calculator, so every bound is chosen to give exact values such as pi over 4, the square root of 2 or the natural log of one plus root 2. Learn the exact values of sine, cosine, tangent and secant at the standard angles, and give your answers in exact form.

Practise it

Corrected exercises: Trigonometric integrals, 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

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Revision sheet: trigonometric integrals (MATH 141)

The essentials

Save one factor from the odd power, convert the rest

No factor to save: lower the degree

The rules in table form

Which factor to save, read off the powers

The mistakes that cost marks

1. Saving the factor from the even power

2. The wrong sign in the half-angle formula

3. Integrating a square with the power rule

4. Replacing the square root by cos x where the cosine is negative

5. Keeping the old lower bound after the substitution u = sec x

6. Losing the one half in the integral of sec cubed

7. Dropping the sign of sin(A - B) when A is smaller than B

8. Copying the csc formula from the sec formula

Which method to choose

Which gesture, by the FORM of the integrand

How the answer is expected to be written

Writing a trigonometric integral for full marks

Check before you hand in

The typical problem, taken apart

One integral, two legitimate substitutions

Learn by heart

Frequently asked questions

How do I know which substitution to use for a power of sine times a power of cosine?

When do I use the half-angle formulas in an integral?

What is the integral of secant cubed and how do I remember it?

Why do two correct methods give different answers for the same trig integral?

Can I use a calculator for trigonometric integrals in MATH 141?

Corrected exercises: Trigonometric integrals, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?