Source: Advanced Calculus I, University of Central Florida

Tags: u-substitution, free response, worked solutions, indefinite integrals, definite integrals, pattern recognition, exam practice, integration techniques

Difficulty: Intermediate to Advanced Prerequisites: Part 1 concepts and Part 2 multiple-choice solutions. Familiarity with trig identities and logarithmic differentiation.

TL;DR

Twenty-two free-response problems ranging from straightforward single-step u-subs to multi-step definite integrals and abstract substitution-and-transformation problems. Every solution below shows the choice of u, the substitution, and the final simplified result. Problems 15, 16, and 17 test conceptual understanding of how substitution interacts with limits and scaling.

Core Content

Problem 12 -- Indefinite Integrals

12(a): ∫ 2x(x² + 1) dx

Let u = x² + 1, du = 2x dx.

∫ u du = u²/2 + C = (x² + 1)² / 2 + C

12(b): ∫ 3t² / (t³ - 4) dt

Let u = t³ - 4, du = 3t² dt.

∫ du/u = ln|u| + C = ln|t³ - 4| + C

12(c): ∫ x√(2x² - 1) dx

Let u = 2x² - 1, du = 4x dx, so x dx = du/4.

∫ √u · (du/4) = (1/4) · (2/3)u^(3/2) + C = u^(3/2)/6 + C = (2x² - 1)^(3/2) / 6 + C

12(d): ∫ 3x · e^(x² + 2) dx

Let u = x² + 2, du = 2x dx, so x dx = du/2.

∫ 3 · e^u · (du/2) = (3/2)e^u + C = (3/2)e^(x² + 2) + C

12(e): ∫ 4x / (x² - 8)³ dx

Let u = x² - 8, du = 2x dx, so x dx = du/2.

∫ 4 · (du/2) / u³ = 2 ∫ u^(-3) du = 2 · u^(-2)/(-2) + C = -1/u² + C = -1/(x² - 8)² + C

12(f): ∫ 2r · e^(3r²) dr

Let u = 3r², du = 6r dr, so r dr = du/6.

∫ 2 · e^u · (du/6) = (1/3)e^u + C = (1/3)e^(3r²) + C

12(g): ∫ 5l² (l³ - 1) dl

Let u = l³ - 1, du = 3l² dl, so l² dl = du/3.

∫ 5u · (du/3) = (5/3) · u²/2 + C = 5u²/6 + C = 5(l³ - 1)² / 6 + C

12(h): ∫ (3x² + 2)√(x³ + 2x) dx

Let u = x³ + 2x, du = (3x² + 2) dx.

∫ √u du = (2/3)u^(3/2) + C = (2/3)(x³ + 2x)^(3/2) + C

12(i): ∫ (6t² + 10t⁴)(t³ + t⁵)^(100) dt

Factor the first part: 6t² + 10t⁴ = 2t²(3 + 5t²). Factor the derivative of t³ + t⁵: d/dt(t³ + t⁵) = 3t² + 5t⁴ = t²(3 + 5t²).

So 6t² + 10t⁴ = 2(3t² + 5t⁴) = 2 du/dt where u = t³ + t⁵.

Let u = t³ + t⁵, du = (3t² + 5t⁴) dt.

∫ 2u^(100) du = 2 · u^(101)/101 + C = 2(t³ + t⁵)^(101) / 101 + C

12(j): ∫ ln³(3x) / (3x) · 3 dx

Rewrite as ∫ ln³(3x) · (1/x) · (1/3) · 3 dx. Actually, let's be more careful. The integrand is ln³(3x) / (3x), and the full expression is ∫ [ln³(3x)]/(3x) · 3 dx. Simplify: 3/(3x) = 1/x.

So the integral is ∫ ln³(3x) · (1/x) dx.

Let u = ln(3x). Then du = (1/x) dx (since d/dx[ln(3x)] = 1/x).

∫ u³ du = u⁴/4 + C = ln⁴(3x) / 4 + C

12(k): ∫ (6x + 5) / (3x² + 5x - 2) dx

Let u = 3x² + 5x - 2, du = (6x + 5) dx.

∫ du/u = ln|u| + C = ln|3x² + 5x - 2| + C

12(l): ∫ (12x + 10) / (9x² + 15x - 6) dx

Factor: 9x² + 15x - 6 = 3(3x² + 5x - 2). Its derivative: d/dx(9x² + 15x - 6) = 18x + 15.

Note that 12x + 10 = (2/3)(18x + 15). Let u = 9x² + 15x - 6, du = (18x + 15) dx.

∫ (2/3) du/u = (2/3) ln|u| + C = (2/3) ln|9x² + 15x - 6| + C

12(m): ∫ cos(3x) / (5 + 2 sin(3x)) dx

Let u = 5 + 2 sin(3x), du = 2 cos(3x) · 3 dx = 6 cos(3x) dx.

So cos(3x) dx = du/6.

∫ (du/6) / u = (1/6) ln|u| + C = (1/6) ln|5 + 2 sin(3x)| + C

12(n): ∫ (2t + 1) · e^(5t² + 5t) dt

Let u = 5t² + 5t, du = (10t + 5) dt = 5(2t + 1) dt.

So (2t + 1) dt = du/5.

∫ e^u · (du/5) = (1/5)e^u + C = (1/5)e^(5t² + 5t) + C

12(o): ∫ sin(ln(ax)) / x dx, where a > 0

Let u = ln(ax). Then du = (1/x) dx.

∫ sin(u) du = -cos(u) + C = -cos(ln(ax)) + C

12(p): ∫ cos³(t) dt

Rewrite using the identity cos²(t) = 1 - sin²(t):

∫ cos(t)(1 - sin²(t)) dt

Let u = sin(t), du = cos(t) dt.

∫ (1 - u²) du = u - u³/3 + C = sin(t) - sin³(t)/3 + C

Problem 13 -- Definite Integrals

13(a): ∫₀¹ x³(1 + x⁴)⁵ dx

Let u = 1 + x⁴, du = 4x³ dx, so x³ dx = du/4.

Limits: x = 0 gives u = 1; x = 1 gives u = 2.

(1/4) ∫₁² u⁵ du = (1/4) · u⁶/6 |₁² = (1/24)(2⁶ - 1⁶) = (1/24)(64 - 1) = 63/24 = 21/8

13(b): ∫_(√(π/4))^(√(2π/3)) x sin(x²) dx

Let u = x², du = 2x dx, so x dx = du/2.

Limits: x = √(π/4) gives u = π/4; x = √(2π/3) gives u = 2π/3.

(1/2) ∫_(π/4)^(2π/3) sin(u) du = (1/2)[-cos(u)] from π/4 to 2π/3

= (1/2)[-cos(2π/3) + cos(π/4)] = (1/2)[1/2 + √2/2] = (1/2)(1 + √2)/2 = (1 + √2)/4

13(c): ∫₋₁³ √(7 + 3x) dx

Let u = 7 + 3x, du = 3 dx, so dx = du/3.

Limits: x = -1 gives u = 4; x = 3 gives u = 16.

(1/3) ∫₄¹⁶ u^(1/2) du = (1/3) · (2/3)u^(3/2) |₄¹⁶ = (2/9)(16^(3/2) - 4^(3/2)) = (2/9)(64 - 8) = (2/9)(56) = 112/9

13(d): ∫₀³ x√(1 + x) dx

Let u = 1 + x, so x = u - 1 and dx = du.

Limits: x = 0 gives u = 1; x = 3 gives u = 4.

∫₁⁴ (u - 1)√u du = ∫₁⁴ (u^(3/2) - u^(1/2)) du

= [2u^(5/2)/5 - 2u^(3/2)/3] from 1 to 4

At u = 4: 2(32)/5 - 2(8)/3 = 64/5 - 16/3 = (192 - 80)/15 = 112/15

At u = 1: 2/5 - 2/3 = (6 - 10)/15 = -4/15

Result: 112/15 - (-4/15) = 116/15

13(e): ∫₀^π cos²(θ/5) sin(θ/5) dθ

Let u = cos(θ/5), du = -sin(θ/5) · (1/5) dθ, so sin(θ/5) dθ = -5 du.

Limits: θ = 0 gives u = cos(0) = 1; θ = π gives u = cos(π/5).

∫ u² · (-5 du) = -5 · u³/3

= (-5/3)[cos³(π/5) - 1] = (5/3)[1 - cos³(π/5)]

cos(π/5) = cos(36°) = (1 + √5)/4... this evaluates to an exact expression. In exact form:

(5/3)(1 - cos³(π/5))

13(f): ∫₀¹ (1 + e^(3x)) / (e^(3x) + 3x) dx

Let u = e^(3x) + 3x. Then du = (3e^(3x) + 3) dx = 3(e^(3x) + 1) dx.

So (1 + e^(3x)) dx = du/3.

∫ (du/3)/u = (1/3) ln|u|

Limits: x = 0 gives u = 1 + 0 = 1; x = 1 gives u = e³ + 3.

Result: (1/3) ln(e³ + 3)

13(g): ∫₀¹ 1/(1 + 9x²) dx

Recognise this as an arctangent form. Write 1 + 9x² = 1 + (3x)², so:

∫ 1/(1 + (3x)²) dx = (1/3) arctan(3x) + C

Evaluate from 0 to 1:

• (1/3) arctan(3) - (1/3) arctan(0) = (1/3) arctan(3)

Since this does not simplify to a standard angle, the answer is (1/3) arctan(3).

If the problem expects a recognisable form: note that arctan(3) is not π/3 or π/4, so the answer stays as (1/3) arctan(3).

Checking the answer choices more carefully: the answer is π/12 only if the upper limit gives arctan(3) = π/4, which it does not. The exact answer is (1/3) arctan(3). However, if the integral were ∫₀^(1/3) 1/(1+9x²) dx, the answer would be π/36. Given the limits 0 to 1, the answer is (1/3) arctan(3).

Problem 14 -- Substitution Applied to Given Identities

14(a)

Given: ∫ dx/(1 + cos x) = tan(x/2) + C. Find ∫ 1/(3 + 3cos(x/4)) dx.

Factor: 3 + 3cos(x/4) = 3(1 + cos(x/4)).

So the integral becomes (1/3) ∫ 1/(1 + cos(x/4)) dx.

Let t = x/4, so x = 4t, dx = 4 dt.

(1/3) ∫ 1/(1 + cos t) · 4 dt = (4/3) · tan(t/2) + C = (4/3) tan(x/8) + C

Answer: (4/3) tan(x/8) + C

14(b)

Given: ∫₀^∞ x/(eˣ + 1) dx = π²/12. Find ∫₀^∞ x/(e^(5x) + 1) dx.

Let u = 5x, so x = u/5 and dx = du/5.

When x = 0, u = 0. When x → ∞, u → ∞.

∫₀^∞ (u/5)/(e^u + 1) · (du/5) = (1/25) ∫₀^∞ u/(e^u + 1) du = (1/25) · π²/12

Answer: π²/300

Problem 15 -- Scaling the Variable Inside the Integral

Given: ∫₀^π ln(a + b cos(x)) dx = 2π ln(2). Find ∫₀^(π/5) ln(a + b cos(5x)) dx.

Let u = 5x, so dx = du/5. When x = 0, u = 0. When x = π/5, u = π.

∫₀^(π/5) ln(a + b cos(5x)) dx = (1/5) ∫₀^π ln(a + b cos(u)) du = (1/5) · 2π ln(2)

Answer: (2π/5) ln(2)

Problem 16 -- Integrals in Terms of K

Given: ∫_a^b f(x) dx = K.

16(a): ∫_(a+5)^(b+5) f(x - 5) dx

Let u = x - 5, du = dx. When x = a + 5, u = a. When x = b + 5, u = b.

∫_a^b f(u) du = K

16(b): ∫_a^b [f(x) + 5] dx

= ∫_a^b f(x) dx + ∫_a^b 5 dx = K + 5(b - a)

Answer: K + 5(b - a)

16(c): ∫_(a/5)^(b/5) f(5x) dx

Let u = 5x, du = 5 dx, so dx = du/5. When x = a/5, u = a. When x = b/5, u = b.

(1/5) ∫_a^b f(u) du = K/5

Problem 17 -- Substitution with Given Integral Value

Given: ∫₃⁶ f(z) dz = 4.

17(a): ∫₁² f(3z) dz

Let u = 3z, du = 3 dz. When z = 1, u = 3. When z = 2, u = 6.

(1/3) ∫₃⁶ f(u) du = (1/3)(4) = 4/3

17(b): ∫_(0.5)² f(7 - 2z) dz

Let u = 7 - 2z, du = -2 dz. When z = 0.5, u = 6. When z = 2, u = 3.

(-1/2) ∫₆³ f(u) du = (-1/2)(−∫₃⁶ f(u) du) = (1/2)(4) = 2

17(c): ∫₄⁷ (f(z - 1) + 5) dz

For the f(z - 1) part, let u = z - 1, du = dz. When z = 4, u = 3. When z = 7, u = 6.

∫₃⁶ f(u) du + ∫₄⁷ 5 dz = 4 + 5(3) = 4 + 15 = 19

Why It Matters / Exam Flags

⚠️ Problems 14, 15, 16, and 17 test whether you understand u-sub at the structural level, not just as a calculation recipe. You are told the value of an integral and asked to find a related one by recognising how substitution transforms limits and scaling. These appear regularly on exams.

⚠️ Problem 12(p) (cos³ t) is a pattern that extends to all odd powers of sin and cos. Learn the identity-based rewrite: peel off one factor, convert the remaining even power using the Pythagorean identity, then substitute.

⚠️ Problem 13(d) (x√(1 + x)) requires rewriting x in terms of u after the substitution. When the "leftover" piece is not a constant but can be expressed in terms of u, the substitution still works. This is a common exam trap.

⚠️ Free-response grading typically awards points for showing the substitution (identifying u and du), correctly rewriting the integral, and the final answer with + C. Skipping steps costs partial credit.

Quick Self-Test

1. Fill in the blank: If u = 3x² + 5x - 2, then du = ______ dx.

1. True or false: ∫ cos³(t) dt can be evaluated by letting u = cos(t).

1. Fill in the blank: If ∫a^b f(x) dx = K, then ∫(a/3)^(b/3) f(3x) dx = ______.

1. True or false: In Problem 12(e), the substitution u = x² - 8 turns the integral into a negative-power-rule problem.

1. Fill in the blank: ∫ (6x + 5)/(3x² + 5x - 2) dx = ______.

Answers: 1. (6x + 5). 2. False; let u = sin(t) instead, so du = cos(t) dt absorbs the peeled-off cos factor. 3. K/3. 4. True. 5. ln|3x² + 5x - 2| + C.

Connections to Other Topics

The abstract substitution problems (15, 16, 17) connect directly to the theory of Riemann sums and how integration interacts with linear transformations of the variable. This idea reappears in probability (changing variables in a PDF), Fourier analysis (scaling and shifting frequencies), and physics (Galilean transformations of integrals over time or space). The trig-power technique in 12(p) extends to reduction formulas, which are a bridge to integration by parts.

Related Terms / Search Tags

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