Source: Advanced Calculus I, University of Central Florida
Tags: u-substitution, pattern recognition, multiple choice, worked solutions, definite integrals, antiderivatives, exam practice
Difficulty: Intermediate Prerequisites: Part 1 concepts (u-sub procedure, pattern recognition, changing limits). Comfort with exponential, logarithmic, and trigonometric integrals.
Eleven multiple-choice problems covering the full range of u-sub patterns: double antiderivatives, expanding before integrating, initial-value problems, exponential and logarithmic integrands, trig compositions, and rational functions requiring algebraic setup before substitution. Each solution below shows the u-choice, the substitution steps, and how to arrive at the correct answer.
Given: f''(x) = 9 cos(3x). Find the most general f(x).
Integrate once:
f'(x) = ∫ 9 cos(3x) dx = 9 · sin(3x)/3 + C₁ = 3 sin(3x) + C₁
Integrate again:
f(x) = ∫ [3 sin(3x) + C₁] dx = 3 · (-cos(3x)/3) + C₁x + D = -cos(3x) + Cx + D
Answer: (B) f(x) = -cos(3x) + Cx + D
Key point: each integration of cos(kx) or sin(kx) introduces a factor of 1/k from the chain rule. Two integrations of 9 cos(3x) give 9/(3·3) = 1 as the coefficient of cos(3x).
Given: Evaluate ∫₀¹ (1 + e^(-x))² dx.
Expand the square first:
(1 + e^(-x))² = 1 + 2e^(-x) + e^(-2x)
Integrate term by term:
∫ 1 dx = x
∫ 2e^(-x) dx = -2e^(-x)
∫ e^(-2x) dx = -e^(-2x)/2
So the antiderivative is x - 2e^(-x) - e^(-2x)/2.
Evaluate from 0 to 1:
At x = 1: 1 - 2e^(-1) - e^(-2)/2
At x = 0: 0 - 2(1) - 1/2 = -5/2
Result: (1 - 2/e - 1/(2e²)) - (-5/2) = 1 + 5/2 - 2/e - 1/(2e²) = 7/2 - 2/e - 1/(2e²)
Answer: (E) 7/2 - 2/e - 1/(2e²)
Key point: not every integral needs u-sub. Sometimes expanding and integrating term by term is cleaner.
Given: f'(x) = 6x·e^(-2x²), f(0) = 1. Find f(-1).
Let u = -2x², so du = -4x dx, meaning x dx = du/(-4).
∫ 6x·e^(-2x²) dx = 6 · ∫ e^u · (du / -4) = -3/2 · e^u + C = -3/2 · e^(-2x²) + C
Apply the initial condition f(0) = 1:
-3/2 · e^(0) + C = 1
-3/2 + C = 1
C = 5/2
So f(x) = -3/2 · e^(-2x²) + 5/2.
Evaluate at x = -1:
f(-1) = -3/2 · e^(-2(1)) + 5/2 = -3/2 · e^(-2) + 5/2 = 5/2 - 3e^(-2)/2
Answer: (A) 5/2 - (3/2)e^(-2)
Given: Evaluate ∫₀¹ (4 - 2x)·e^(8x - 2x²) dx.
Notice the exponent: g(x) = 8x - 2x². Its derivative is g'(x) = 8 - 4x = 2(4 - 2x).
So (4 - 2x) dx = g'(x)/2 dx = du/2 where u = 8x - 2x².
∫ (4 - 2x)·e^(8x - 2x²) dx = (1/2) ∫ e^u du = e^u / 2 = e^(8x - 2x²) / 2
Change limits:
x = 0: u = 0
x = 1: u = 8 - 2 = 6
Result: (1/2)(e^6 - e^0) = (e^6 - 1)/2
Answer: (A) (1/2)(e⁶ - 1)
Given: Evaluate ∫₀^(π/4) (2e^(tan x) + 5) / cos²(x) dx.
Since 1/cos²(x) = sec²(x) and d/dx[tan x] = sec²(x), let u = tan x, du = sec²(x) dx.
The integrand becomes (2e^u + 5) du.
∫ (2e^u + 5) du = 2e^u + 5u
Change limits:
x = 0: u = tan(0) = 0
x = π/4: u = tan(π/4) = 1
Result: (2e^1 + 5·1) - (2e^0 + 0) = 2e + 5 - 2 = 2e + 3
Answer: (A) 2e + 3
Given: ∫ (4/x)(1 + 2 ln x)³ dx.
Let u = 1 + 2 ln x, so du = (2/x) dx, meaning (1/x) dx = du/2.
∫ 4 · u³ · (du/2) = 2 ∫ u³ du = 2 · u⁴/4 = u⁴/2 = (1 + 2 ln x)⁴ / 2 + C
Answer: (B) (1/2)(1 + 2 ln x)⁴ + C
Given: Evaluate ∫₁ᵉ (1/x)(f'(ln x) + 2) dx, where f(0) = 1 and f(1) = 4.
Let u = ln x, so du = (1/x) dx.
When x = 1, u = 0. When x = e, u = 1.
∫₀¹ (f'(u) + 2) du = [f(u) + 2u] from 0 to 1
= (f(1) + 2) - (f(0) + 0) = (4 + 2) - 1 = 5
Answer: (B) 5
Key point: the Fundamental Theorem of Calculus says ∫ f'(u) du = f(u). The given endpoint values of f are exactly what you need.
Given: Evaluate ∫₀¹ 6x / (1 + x²) dx.
Let u = 1 + x², so du = 2x dx, meaning x dx = du/2.
∫ 6 · (du/2) / u = 3 ∫ du/u = 3 ln|u|
Change limits:
x = 0: u = 1
x = 1: u = 2
Result: 3 ln 2 - 3 ln 1 = 3 ln 2
Answer: (D) 3 ln 2
Given: Evaluate ∫_(π/4)^(3π/4) (6 cos x - 2 sin x) / (6 sin x + 2 cos x) dx.
Let u = 6 sin x + 2 cos x, so du = (6 cos x - 2 sin x) dx. The numerator is exactly du.
∫ du/u = ln|u|
Change limits:
x = π/4: u = 6(√2/2) + 2(√2/2) = 3√2 + √2 = 4√2
x = 3π/4: u = 6(√2/2) + 2(-√2/2) = 3√2 - √2 = 2√2
Result: ln(2√2) - ln(4√2) = ln(2√2 / 4√2) = ln(1/2) = -ln 2
Answer: (B) -ln 2
Given: Evaluate ∫₀¹ (x² + 4x + 1) / (3x² + 3) dx.
Split the fraction. Note 3x² + 3 = 3(x² + 1).
Divide: (x² + 4x + 1) / (3(x² + 1)) = (x² + 1 + 4x) / (3(x² + 1)) = 1/3 + (4x) / (3(x² + 1))
So the integral becomes:
∫₀¹ [1/3 + (4x)/(3(x² + 1))] dx
First piece: ∫₀¹ 1/3 dx = 1/3.
Second piece: (4/3) ∫₀¹ x/(x² + 1) dx. Let u = x² + 1, du = 2x dx.
(4/3)(1/2) ∫₁² du/u = (2/3) ln|u| from 1 to 2 = (2/3) ln 2
Total: 1/3 + (2/3) ln 2 = (1 + 2 ln 2) / 3
Answer: (B) (1 + 2 ln 2) / 3
Given: Evaluate ∫_e^(e⁴) 5 / (x√(ln x)) dx.
Let u = ln x, so du = (1/x) dx.
∫ 5 / √u · du = 5 ∫ u^(-1/2) du = 5 · 2u^(1/2) = 10√u
Change limits:
x = e: u = ln e = 1
x = e⁴: u = ln(e⁴) = 4
Result: 10√4 - 10√1 = 20 - 10 = 10
Answer: (E) 10
⚠️ Problems 2 and 10 show that algebraic simplification (expanding, splitting fractions) often comes before any substitution. Do not force a u-sub when algebra can simplify first.
⚠️ Problem 7 tests whether you understand the Fundamental Theorem at a conceptual level, not just as a calculation. Recognising ∫ f'(u) du = f(u) is critical.
⚠️ Problem 9 is the classic "numerator is the derivative of the denominator" pattern. Train yourself to check this immediately whenever you see a fraction.
⚠️ Watch your chain-rule constants. In Problem 1, two successive integrations of cos(3x) produce a factor of 1/9, which exactly cancels the leading 9. Under time pressure students often lose track of these.
Q: In Problem 4, how do you recognise that (4 - 2x) is related to the derivative of the exponent 8x - 2x²?
A: Differentiate the exponent: d/dx(8x - 2x²) = 8 - 4x = 2(4 - 2x). The integrand factor (4 - 2x) is half the derivative, so the integral picks up a factor of 1/2.
Q: Why does expanding work better than u-sub in Problem 2?
A: There is no single inner function whose derivative accounts for the rest of the integrand. The expression (1 + e^(-x))² is a sum, not a composition with a convenient du sitting alongside. Expanding and integrating term by term is straightforward.
Q: In Problem 9, what if the numerator were not exactly the derivative of the denominator?
A: If it were off by a constant multiple, you would adjust. If it were a genuinely different expression, the ln|u| pattern would not apply and you would need a different method entirely.
Q: How do you handle the absolute value in ln|u| for Problem 8?
A: Since u = 1 + x² is always positive for real x, |u| = u and you can drop the absolute value. Always check the sign of your u on the interval of integration.
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