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CEGEP Calculus II — Techniques of Integration Worksheet

By Marius — private math tutor in Montreal. Updated 30 August 2026

By this point in Calculus II you have met every standard technique separately, and the hard part is no longer executing one — it is looking at an unfamiliar integrand and knowing, before you write anything, which technique it wants. This set works through integration by parts and repeated parts, trigonometric integrals, trigonometric substitution and partial fractions, and then spends a whole sheet on nothing but the choice itself. Have a look on this page, then print the free PDF when you want to write on it.

Practice worksheet — free PDF

6 pages 7 questions Letter size, print-ready

Download the worksheet Read the questions here

No email, no account, no watermark. Teachers: photocopy it for your classes freely. Worked solutions and 6 harder problems come with the CEGEP Calculus II bundle.

All 7 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

Q1Integration by Parts

Evaluate each integral by parts, stating your choice of $u$ and $d v$ before you begin.
1. $\int x e^{- x} d x$
2. $\int_{1}^{e} x^{2} \ln x d x$
Q2Repeated Integration by Parts

Let $F (x) = \int x^{2} \cos x d x$ .
1. Find $F (x)$ , applying integration by parts twice.
2. Use your antiderivative to evaluate $\int_{0}^{π / 2} x^{2} \cos x d x$ .
Q3Trigonometric Integrals

Evaluate, using an appropriate identity in each case.
1. $\int \sin^{3} x \cos^{2} x d x$
2. $\int_{0}^{π / 4} \sin^{2} (2 x) d x$
State which feature of the integrand told you which identity to reach for.
Q4Trigonometric Substitution

Evaluate each integral by a trigonometric substitution. State the substitution, and describe the right triangle you use to return to $x$ .
1. $\int \sqrt{16 - x^{2}} d x$ , for $- 4 < x < 4$
2. $\int \frac{d x}{(x^{2} + 9)^{3 / 2}}$
Q5Partial Fractions

Decompose each integrand into partial fractions, then integrate.
1. $\int \frac{5 x - 1}{x^{2} - x - 2} d x$
2. $\int \frac{x + 5}{x (x + 1)^{2}} d x$
Q6Choosing an Integration Technique

For each integral below, name the technique you would use — ordinary substitution, integration by parts, a trigonometric identity, trigonometric substitution, or partial fractions — give a one-sentence reason drawn from the form of the integrand, and write down the first line only: the substitution, the choice of $u$ and $d v$ , or the shape of the decomposition. Do not evaluate any of them.
1. $\int \frac{x}{x^{2} + 6 x + 13} d x$
2. $\int x^{3} \ln x d x$
3. $\int \cos^{5} x d x$
4. $\int \frac{d x}{x^{2} \sqrt{25 - x^{2}}}$
5. $\int \frac{3 x - 2}{x^{2} - 4 x + 3} d x$
6. $\int x \sqrt{x^{2} - 7} d x$
Q7Synthesis — drawing on several topics in this set

Evaluate $\int_{0}^{1} x \ln (x + 1) d x$ exactly.
1. Carry the computation through.
2. Name the two techniques from this set that the problem required, and state the step at which each became necessary.

The 6 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

Which stream is this for? This sheet is built for CEGEP Calculus II — the course numbered 201-NYB-05 under the legacy system and 201-SN3-RE under the current one, in the Sciences de la nature programme, under competency 0M03. Two things students expect here are deliberately elsewhere. Numerical integration — the trapezoidal rule, Simpson's rule — is not part of 0M03 and is not on this sheet. Improper integrals are part of the course but belong with the indeterminate forms that go with them, so they have their own worksheet in this series; every integral here is proper and every antiderivative is one you can write down exactly.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

The mark most people lose is for choosing, not for grinding

The competency asks for the pertinent application of the techniques of integration, and that word carries the marks. A student who can execute parts, a trigonometric substitution and a partial-fraction decomposition on demand can still lose an exam question by spending twenty minutes pushing the wrong one. Q6 is built on exactly this: six integrands, and for each you name the technique, give a one-sentence reason drawn from the form of the integrand, and write the first line only — nothing is evaluated. Practise it out loud, because "partial fractions, because it is a proper rational function whose denominator factors into distinct linear factors" is an answer, and "I tried substitution and it didn't work" is not.

Which technique, and what tells you

Ask these in order, of the integrand as written, and stop at the first yes.

1
Does part of the integrand supply the derivative of another part?
Then it is an ordinary substitution, and none of the heavier machinery is needed. This test comes first because it overrides every cue below: a radical, a quadratic denominator or a trigonometric power that would otherwise call for something elaborate is settled in one line when the differential is already sitting there. Q6(f) is placed there to catch you.
2
Is it a product of two unrelated kinds of function?
Then it is integration by parts. The cue is a product in which one factor gets simpler when differentiated and the other is one you can integrate — a polynomial with an exponential or with a sine or cosine, or anything at all multiplied by a logarithm or an inverse trigonometric function. Those last two are the strongest signal in the list, because a logarithm is a function you can differentiate but not readily integrate, so it has to be the u.
3
Is it a rational function — a polynomial over a polynomial?
Then it is partial fractions, but only after two checks. Is it proper, with numerator degree strictly below the denominator's? If not, divide first. And does the denominator actually factor? An irreducible quadratic denominator with nothing to substitute is not a partial-fractions problem at all — it is completing the square and an arctangent, which is where Q6(a) lands.
4
Is everything trigonometric, in powers?
Then it is a trigonometric identity, and the parity of the exponents picks which one. See the section below.
5
Is there a radical of the form √(a² − x²), √(x² + a²) or √(x² − a²) and nothing to absorb it?
Then it is a trigonometric substitution — the technique of last resort, because it costs the most work. "Nothing to absorb it" is the load-bearing clause: you only reach step 5 because step 1 said no.

The three near-misses that decide the question. Each is a pair of integrands that look alike and want different techniques, and each of them appears somewhere on this sheet.
A radical with a factor of x beside it, versus without. A radical does not by itself force a trigonometric substitution. If the rest of the integrand supplies the derivative of what sits under the root, an ordinary substitution finishes the job in a line; if there is nothing to absorb, the radical has to be cleared by an identity instead. Ask what the numerator supplies, not what the denominator looks like.
A factorable quadratic denominator, versus an irreducible one. Partial fractions needs real factors to split into. Check the discriminant before you write a single A ⁄ (x − r): if it is negative there are none, and the route is completing the square, aiming at ∫ du ⁄ (u² + a²) and an arctangent.
A proper fraction, versus an improper one. No decomposition can start on a fraction whose numerator has degree at least the denominator's. Long division comes first, always — and Q7 is built so that the first step leaves you an improper fraction to notice.

Integration by parts: choosing u, and knowing when to stop

Q1 asks you to state your choice of u and dv before you begin, which is the habit worth building — the whole difficulty of parts sits in that one decision.

∫ u dv = uv − ∫ v du

The choice is judged against one criterion: the new integral ∫ v du must be easier than the one you started with. So u should be the factor that gets simpler when differentiated, and dv the factor you can actually integrate. A logarithm or an inverse trigonometric function is nearly always u, because it has no elementary antiderivative you would want to reach for but a very simple derivative. A polynomial is u when the other factor is an exponential or a sine or cosine, because differentiating drops its degree while the other factor is no harder than before. Choose the other way round and you still get a valid identity, just a worse integral than you began with; that is not an error to patch with algebra, it is the signal to go back and swap.

Repeated parts terminates when the polynomial does. Q2 needs two passes because each pass differentiates the polynomial factor once, dropping its degree by one — so a polynomial factor of degree n needs n passes, and then the polynomial is gone. Knowing the count in advance is what stops you losing the thread halfway: give the inner integral its own line, evaluate it completely, and only then substitute it back. Q2 also splits the work the way an exam does — find the general antiderivative first, then use it to evaluate the definite integral, rather than dragging limits through both passes.

Trigonometric integrals: the parity of the exponents chooses the identity

Q3 asks you to state which feature of the integrand told you which identity to reach for, and for powers of sine and cosine one rule answers it every time.

Some exponent odd. Peel one factor off the odd power to serve as the differential, convert what remains with sin²x + cos²x = 1, and substitute. The odd power is what makes this work — it leaves an even number of factors behind, and an even power is exactly what the Pythagorean identity can rewrite.
All exponents even. Nothing can be peeled off without stranding an odd power, so no substitution starts. Use a half-angle identity — sin²θ = (1 − cos 2θ) ⁄ 2, cos²θ = (1 + cos 2θ) ⁄ 2 — which sets up no substitution but lowers the powers, and repeat if what comes out is still even. The identity applies to whatever sits inside the function, so a squared sine of 2x becomes an expression in cos 4x, and that inner factor of 2 carries through into the antiderivative.

Trigonometric substitution: the form of the radical, and the trip back

Q4 asks for the substitution and for a description of the right triangle you use to return to x. Both halves are marked, and the second is the one that gets skipped.

Matching the radical to the substitution

Each identity is chosen to make the radical come out as a single trigonometric function with no root left.

1
√(a² − x²) → x = a sin θ
Because a² − a² sin²θ = a² cos²θ. Take θ in [−π ⁄ 2, π ⁄ 2], where cos θ ≥ 0, so the root is a cos θ and no absolute value is needed.
2
√(x² + a²) → x = a tan θ
Because a² tan²θ + a² = a² sec²θ. Take θ in (−π ⁄ 2, π ⁄ 2), where sec θ > 0. The same substitution handles a power of (x² + a²) with no root written at all — a three-halves power in a denominator, for instance.
3
√(x² − a²) → x = a sec θ
Because a² sec²θ − a² = a² tan²θ. Restrict θ to the branch matching the interval you were given, so that tan θ ≥ 0 and the root stays positive there.
4
Then draw the triangle
Your antiderivative is in θ and the question was in x, so you have to convert back. Label a right triangle straight from the substitution — if sin θ = x ⁄ a then x is the opposite side and a the hypotenuse — and read every other ratio off it. That is how sin θ, cos θ, tan θ and sec θ turn back into expressions in x without memorising a single inverse-function identity.

Two habits keep this technique honest. State the range of θ when you state the substitution, because that is what licenses writing the radical as a positive multiple of a trigonometric function rather than an absolute value. And differentiate the result: a trigonometric substitution is long enough that an early slip survives all the way to the end, and one line of differentiation catches it.

Partial fractions: the shape of the decomposition is dictated, not guessed

The two integrands in Q5 are there to make one point — how the denominator factors tells you the form of the decomposition completely, and the only work left is finding the constants.

Distinct linear factors: one term with a constant numerator for each factor.
A repeated linear factor: one term for every power up to the multiplicity — a squared factor contributes a term over the factor and a term over its square, not one term. Omitting the lower power is the commonest structural error here, and it shows up as a system of equations with no solution.
An irreducible quadratic factor: its numerator must be a general polynomial of degree one, Bx + D, never a constant. A constant offers one free parameter where the identity needs two, so the coefficients cannot be matched at all.

For the constants, substituting the roots of the linear factors kills most of the terms and gives them almost for free; whatever is left comes from comparing the coefficients of a single power. Then two things are worth writing down. A term sitting over a repeated factor integrates by the power rule, not to a logarithm — only a first power of a linear factor gives a log. And the domain restrictions are part of the answer: the antiderivative is valid on an interval that excludes the zeros of the denominator, so name them.

The synthesis question, and what it is really testing

Q7 asks you to carry a computation through and then name the two techniques the problem required and the step at which each became necessary. That second part is not a formality. Real integrals rarely announce one technique; they need one to get started and another to finish what the first left behind — a parts step hands you back a rational function, a trigonometric substitution hands you back a trigonometric integral. Being able to say where the switch happened is the difference between having solved the problem and having copied a procedure, and it is what makes the next unfamiliar integral tractable.

Getting the most out of it

Write the first line before you commit to anything

For every integral here, write down the technique, the reason drawn from the form of the integrand, and the opening line — the substitution, the choice of u and dv, or the shape of the decomposition. Then look at that line and ask whether the integral it produces is genuinely easier than the one you had. Thirty seconds there saves the twenty minutes a wrong technique costs, and it is the exact skill Q6 drills.

Check the fraction before you decompose it

Two questions, always in this order, before a single constant is named: is the fraction proper, and does the denominator factor over the reals? A no to the first means long division first; a no to the second means completing the square, not partial fractions. Both checks take seconds, and both failures stay invisible until the algebra collapses several lines later.

Differentiate every antiderivative you produce

Each of these techniques is long enough to hide a sign error or a dropped factor, and every answer on this sheet can be verified in one line by differentiating it back to the integrand. Do that before you look anything up. On a definite integral, add a check on the size as well: if the integrand is positive across the interval and your value is negative, the slip is upstream of the arithmetic.

Want the solutions, or something more challenging?

The worksheet, the questions above and every explanation on this page stay free permanently. Three more PDFs exist for this topic — the worked answer key, a harder problem set, and the answer key to that. They come with the CEGEP Calculus II Solutions Bundle, which is what keeps the rest of the series free.

What else exists for Techniques of Integration

Three PDFs · 14 pages · all three are in the bundle below.

Answer key — 4 pages. All 7 questions worked step by step, including the restrictions and the justifications. Not a list of final answers.
Challenge problems — 6 pages, 6 problems. A separate sheet at exam-plus difficulty covering the same 6 concepts. Harder than anything on the free sheet.
Challenge answer key — 4 pages. Every challenge problem worked to the same standard, with the checks shown.
PDF, letter size, print-ready.

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Common questions

Is this worksheet really free?

Yes — the questions are on this page to read, and the PDF downloads directly, no email and no account. The one paid item is optional: the complete CEGEP Calculus II Solutions Bundle, which covers every set at this level.

Is this for 201-NYB-05 or 201-SN3-RE?

Both — they are the same course under two numbering systems, the second being the current one. Whichever number is on your timetable, the techniques of integration sit in the same block of competency 0M03, and this sheet is built against that.

How do I tell whether an integral needs a trigonometric substitution?

Look at what the rest of the integrand supplies, not at the radical. If the factor outside the root is a constant multiple of the derivative of what is under it, an ordinary substitution settles the integral in a line and a trigonometric substitution would be wasted work. Only when there is nothing to absorb does the radical have to be cleared by an identity.

When can I not use partial fractions on a rational function?

In two situations, and they are checked separately. If the fraction is improper — the numerator's degree is at least the denominator's — no decomposition can start, and you divide first. And if the denominator is an irreducible quadratic there are no real linear factors to split into at all: check the discriminant, and if it is negative, complete the square and expect an arctangent.

Which factor should be u in integration by parts?

The one that gets simpler when you differentiate it, provided the other is something you can integrate. A logarithm or an inverse trigonometric function is almost always u, because differentiating it is easy and integrating it is not. If the integral you are left with looks worse than the one you started with, the choice was backwards — swap and start again rather than pressing on.

Are Simpson's rule and the trapezoidal rule on this sheet?

No. Numerical integration is not part of competency 0M03, so it is left out rather than included as filler. Every integral on this sheet is one you can evaluate exactly.

Where are the improper integrals?

On their own worksheet in this series, alongside the indeterminate forms and L'Hospital's rule, since deciding whether an improper integral converges leans on the same limit reasoning. They are very much part of the course; they are simply a different sheet. Every integral here has finite limits and a bounded integrand.

Can teachers use this in class?

Yes. Print and photocopy it for your own classes freely — I just ask that the tutorinmontreal.ca footer stays on the page.

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