Calculus II in CEGEP: how to choose the right integration technique

The real challenge in Calculus II: picking the technique

In Calculus II (Integral Calculus, 201-NYB), the course that follows Calculus I in the Science program, most students can apply each technique on its own. The trouble starts on the exam: faced with an integral and no hint, which method do you use? According to the Cégep à distance course description, the course covers integration techniques, improper integrals, and sequences and series, among other topics. This guide gives you a way to decide, with worked examples checked by differentiating every answer.

Rule zero, before any technique: simplify. Expand a product, split a fraction into separate terms, apply a trig identity. Plenty of "hard" integrals turn into basic formulas after one line of algebra.

The decision chart

Treat the chart as a starting point, not a law. If a method hasn't made the integral simpler after two or three lines, back up and try the next one.

Substitution: your first reflex

Look for an "inside" expression whose derivative also shows up in the integral. In ∫ 2x cos(x^2) dx, let u = x^2, so du = 2x dx, and the integral becomes ∫ cos u du = sin u + C = sin(x^2) + C. Check: the derivative of sin(x^2) is cos(x^2) · 2x.

The derivative only needs to be there up to a constant. For ∫ x/(x^2 + 1) dx, let u = x^2 + 1, du = 2x dx, so x dx = du/2, giving ½ ln(x^2 + 1) + C. With a definite integral, remember to change the bounds too.

Integration by parts and the LIATE heuristic

The formula ∫ u dv = uv − ∫ v du is mainly for a product of two different kinds of functions. To pick u, the LIATE heuristic suggests this order: Logarithmic, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential. Whichever comes first becomes u, and the rest is dv.

• ∫ x e^x dx: u = x, dv = e^x dx, so ∫ x e^x dx = x e^x − ∫ e^x dx = (x − 1)e^x + C.
• ∫ ln x dx: u = ln x, dv = dx, so x ln x − ∫ x · (1/x) dx = x ln x − x + C.
• ∫ x ln x dx: u = ln x (L beats A), dv = x dx, which gives (x^2/2) ln x − x^2/4 + C.
• ∫ e^x sin x dx: after integrating by parts twice, the original integral reappears. Solve for it algebraically: e^x (sin x − cos x)/2 + C.

LIATE is a heuristic, not a rule. For ∫ x^3 e^(x^2) dx, take dv = x e^(x^2) dx (which you can integrate by substitution) and u = x^2, giving (x^2 − 1)e^(x^2)/2 + C.

Trig integrals and trig substitution

With powers of sine and cosine, identities do the heavy lifting. If one power is odd, set one factor aside: ∫ sin^3 x dx = ∫ (1 − cos^2 x) sin x dx, then u = cos x gives −cos x + cos^3 x / 3 + C. If every power is even, use the half-angle identity sin^2 x = (1 − cos 2x)/2, so ∫ sin^2 x dx = (x − sin x cos x)/2 + C.

Trig substitution targets square roots of quadratics:

• √(a^2 − x^2): x = a sin θ
• √(a^2 + x^2): x = a tan θ
• √(x^2 − a^2): x = a sec θ

Example: for ∫ dx / (x^2 √(4 − x^2)), let x = 2 sin θ. The integral becomes ¼ ∫ csc^2 θ dθ = −¼ cot θ + C. A right triangle gives cot θ = √(4 − x^2)/x, so the answer is −√(4 − x^2)/(4x) + C. Careful though: ∫ x/√(4 − x^2) dx only needs the plain substitution u = 4 − x^2. Always check the simplest route first.

Partial fractions and improper integrals

Partial fractions

For a ratio of polynomials, compare degrees first: if the numerator's degree is greater than or equal to the denominator's, do long division. Then factor the denominator. Example: (5x − 1)/((x − 1)(x + 1)) = A/(x − 1) + B/(x + 1), so 5x − 1 = A(x + 1) + B(x − 1). Setting x = 1 gives A = 2; setting x = −1 gives B = 3. Result: 2 ln|x − 1| + 3 ln|x + 1| + C. Repeated factors and irreducible quadratic factors need extra terms.

Improper integrals

An infinite bound or a vertical asymptote on the interval means you need a limit. ∫ from 1 to ∞ of 1/x^2 dx = lim (1 − 1/b) = 1 (converges), while ∫ from 1 to ∞ of 1/x dx diverges. In general, ∫ from 1 to ∞ of 1/x^p dx converges only when p > 1, and ∫ from 0 to 1 of 1/x^p dx only when p < 1 (for instance, ∫ from 0 to 1 of 1/√x dx = 2). Classic trap: ∫ from −1 to 1 of 1/x^2 dx is not −2. It diverges, because 1/x^2 is undefined at 0.

Series: which convergence test should you use?

1. Divergence test. If the terms don't go to 0, the series diverges: Σ n/(n + 1) diverges because n/(n + 1) goes to 1. The converse is false: the harmonic series Σ 1/n diverges even though 1/n goes to 0.
2. Known series. A geometric series Σ ar^n converges when |r| < 1; a p-series Σ 1/n^p converges when p > 1.
3. Comparison. Σ 1/(n^2 + 1) converges because 1/(n^2 + 1) < 1/n^2. The limit comparison test helps when the inequality is awkward to prove.
4. Ratio test. Ideal with factorials or exponentials: for Σ 2^n/n!, the ratio is 2/(n + 1), which goes to 0, so the series converges. If the limit is 1, the test tells you nothing.
5. Alternating series. Σ (−1)^(n+1)/n converges because 1/n decreases to 0, but only conditionally, since Σ 1/n diverges.
6. Integral test. For a positive, decreasing function you can integrate, such as Σ 1/(n ln n), which diverges.

How to practise for the exam

• Mix your problems. Textbook chapters tell you the technique in advance; the exam won't. Work through mixed sets and name the technique before you calculate anything.
• Differentiate your answer. It's the most reliable way to check an antiderivative, and it often takes under a minute.
• Brush up on algebra and trig. Factoring, completing the square and identities come up everywhere.
• Keep an error log. Write down every wrong turn and the clue that should have pointed you elsewhere.

If a technique keeps tripping you up, our Calculus II tutoring page explains how Stellaire Académie can help, and you can book a session. To shore up the previous course, see common Calculus 1 mistakes, and plan the end of term with our final exam grade calculator.

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