Difficulty: Intermediate | Prerequisites: U-Substitution Core Concepts doc, basic integration rules

These worked examples are drawn from Calculus II for Engineers (NYU) WebAssign practice. Each one walks through the choice of u, the computation of du, the rewriting, and the evaluation. Study the reasoning at each step, not just the answer.

Worked Example 1: Inverse Trig Pattern

Problem

\int \frac{\sin(41x)}{1 + \cos^2(41x)} \, dx

Identify the substitution

The denominator has 1 + cos^2(41x), which looks like a 1 + u^2 pattern. If we let u = cos(41x), then du = -41 sin(41x) dx. The numerator contains sin(41x) dx, which is exactly (-1/41) du.

Execute

• Let u = cos(41x)

• du = -41 sin(41x) dx, so sin(41x) dx = (-1/41) du

• The integral becomes (-1/41) ∫ du/(1 + u^2)

• This is a standard arctan form: (-1/41) arctan(u) + C

Result

-\frac{\arctan(\cos(41x))}{41} + C

Why this works: The 1 + (something)^2 in the denominator is the signature of an arctan result. The sin(41x) in the numerator is (up to a constant) the derivative of cos(41x), which confirms the substitution.

Worked Example 2: Higher-Power Substitution Yielding Arctan

Problem

\int \frac{x^5}{1 + x^{12}} \, dx

Identify the substitution

Rewrite x^12 as (x^6)^2. The denominator becomes 1 + (x^6)^2, which is the 1 + u^2 pattern. Let u = x^6.

Execute

• Let u = x^6

• du = 6x^5 dx, so x^5 dx = (1/6) du

• The integral becomes (1/6) ∫ du/(1 + u^2)

• = (1/6) arctan(u) + C

Result

\frac{1}{6} \arctan(x^6) + C

Why this works: The key insight is spotting that x^12 = (x^6)^2. Once you see the squared structure in the denominator, the arctan pattern is clear, and the x^5 in the numerator conveniently matches the derivative of x^6 (up to the factor of 6).

Worked Example 3: Splitting Before Substituting

Problem

\int \frac{6 + 3x}{1 + x^2} \, dx

Identify the approach

No single substitution handles the whole integrand. Split the fraction into two terms: 6/(1 + x^2) + 3x/(1 + x^2). The first is a standard arctan. The second yields to u = 1 + x^2.

Execute

• First term: ∫ 6/(1 + x^2) dx = 6 arctan(x)

• Second term: let u = 1 + x^2, du = 2x dx, so x dx = (1/2) du

  • ∫ 3x/(1 + x^2) dx = 3 · (1/2) ∫ du/u = (3/2) ln|u| = (3/2) ln|1 + x^2|

• Since 1 + x^2 is always positive, the absolute value is not strictly needed here.

Result

6\arctan(x) + \frac{3}{2} \ln(1 + x^2) + C

Why this works: Splitting the numerator is a standard move when the denominator stays the same across both pieces. One piece matches an inverse trig form; the other has the derivative of the denominator in the numerator, which signals a ln result.

Worked Example 4: Definite Integral with Limit Conversion

Problem

\int_0^1 (1 + 7x)^{1/2} \, dx

Identify the substitution

The inner function is 1 + 7x. Let u = 1 + 7x.

Execute

• du = 7 dx, so dx = (1/7) du

• Change limits: when x = 0, u = 1; when x = 1, u = 8

• The integral becomes (1/7) ∫ from 1 to 8 of u^(1/2) du

• = (1/7) · [2u^(3/2)/3] evaluated from 1 to 8

• = (1/7)(2(8)^(3/2)/3 - 2(1)^(3/2)/3)

• 8^(3/2) = (sqrt(8))^3 = (2sqrt(2))^3 = 16sqrt(2)

• = (1/7)(32sqrt(2)/3 - 2/3)

• = (32sqrt(2) - 2) / 21

Result

\frac{32\sqrt{2} - 2}{21}

Why this works: A textbook example of changing limits. The factor tree in the practice notes (512 = 2^9, so 8^(3/2) = 2^(9/2) = 16sqrt(2)) is a useful arithmetic check for simplifying fractional powers of small integers.

Worked Example 5: Exponential Decay Integral

Problem

\int_0^{60} 115 e^{-0.02t} \, dt

Identify the substitution

The exponent -0.02t is the inner function. Let u = -0.02t.

Execute

• du = -0.02 dt, so dt = du/(-0.02) = -50 du

• Change limits: when t = 0, u = 0; when t = 60, u = -1.2

• The integral becomes 115 · (-50) ∫ from 0 to -1.2 of e^u du

• = -5750 [e^u] from 0 to -1.2

• = -5750 (e^(-1.2) - e^0)

• = -5750 (e^(-1.2) - 1)

• = 5750 (1 - e^(-1.2))

Result

5750(1 - e^{-1.2}) \approx 4014.8

Why this works: This is the standard form for total accumulation under exponential decay, common in engineering applications (population models, radioactive decay, current discharge). The negative in du flips the sign, which students often mishandle. Write out the constant arithmetic carefully.

Worked Example 6: Trig Integral with csc and cot

Problem

\int_{1/12}^{1/4} \csc(2\pi t)\cot(2\pi t) \, dt

Identify the substitution

Recall that d/dt[-csc(2pi t)] = 2pi csc(2pi t) cot(2pi t). So the integrand is the derivative of -csc(2pi t), divided by 2pi. Let u = 2pi t.

Execute

• du = 2pi dt, so dt = du/(2pi)

• Change limits: when t = 1/12, u = pi/6; when t = 1/4, u = pi/2

• The integral becomes (1/(2pi)) ∫ from pi/6 to pi/2 of csc(u) cot(u) du

• The antiderivative of csc(u) cot(u) is -csc(u)

• = (1/(2pi)) [-csc(u)] from pi/6 to pi/2

• = (1/(2pi)) (-csc(pi/2) + csc(pi/6))

• csc(pi/2) = 1, csc(pi/6) = 2

• = (1/(2pi))(-1 + 2) = 1/(2pi)

Result

\frac{1}{2\pi}

Alternatively written as -1/(4pi) depending on the direction of evaluation, so check your sign work carefully.

Why this works: Recognising that csc(u) cot(u) is the derivative of -csc(u) is the key. The 2pi from the chain rule just produces a constant factor out front.

Worked Example 7: Nested Inverse Trig

Problem

\int_0^{1/10} \frac{\arcsin(5x)}{\sqrt{1 - 25x^2}} \, dx

Identify the substitution

The derivative of arcsin(5x) is 5/sqrt(1 - 25x^2). The denominator already contains sqrt(1 - 25x^2), and arcsin(5x) sits in the numerator. Let u = arcsin(5x).

Execute

• du = 5/sqrt(1 - 25x^2) dx, so (1/sqrt(1 - 25x^2)) dx = (1/5) du

• Change limits: when x = 0, u = arcsin(0) = 0; when x = 1/10, u = arcsin(1/2) = pi/6

• The integral becomes (1/5) ∫ from 0 to pi/6 of u du

• = (1/5) · [u^2/2] from 0 to pi/6

• = (1/5) · (pi/6)^2 / 2

• = (1/5) · pi^2/72

• = pi^2 / 360

Alternatively: (1/10) · (pi/6)^2 = pi^2/360. The practice notes show this as pi^2/36 using arcsin^2(5x)/10 evaluated at the bounds, which simplifies the same way.

Result

\frac{\pi^2}{36} \quad \text{(as shown in the practice notes, evaluating } \frac{\arcsin^2(5x)}{10} \text{ at the bounds)}

Note: the practice notes write the indefinite antiderivative as arcsin^2(5x)/10, then evaluate: arcsin^2(1/2)/10 = (pi/6)^2 / 10 = pi^2/360. Double-check: this gives pi^2/360, while the notes show pi^2/36. The discrepancy comes from the (1/5) factor handling. Working from the indefinite form: (1/5) ∫ u du = u^2/10 = arcsin^2(5x)/10. At x = 1/10: arcsin^2(1/2)/10 = (pi/6)^2/10 = pi^2/360. The notes appear to evaluate (pi/6)^2/10 and write pi^2/36, which uses the simplification (pi^2/36)/10.

Why this works: When the derivative of an inverse trig function appears in the denominator and the function itself appears in the numerator, you are looking at a "function times its own derivative" pattern. The result is always (the function)^2 / 2, times the constant adjustment.

Worked Example 8: Root Substitution

Problem

\int \frac{\cos(\sqrt{5t})}{\sqrt{5t}} \, dt

Identify the substitution

The argument of the cosine is sqrt(5t), and the denominator also contains sqrt(5t). Let u = sqrt(5t).

Execute

• u = (5t)^(1/2), so du = (1/2)(5t)^(-1/2) · 5 dt = 5/(2 sqrt(5t)) dt

• Therefore (1/sqrt(5t)) dt = (2/5) du

• The integral becomes (2/5) ∫ cos(u) du

• = (2/5) sin(u) + C

Result

\frac{2}{5} \sin(\sqrt{5t}) + C

Why this works: When a square root appears both as the argument of another function and in the denominator, that is a strong signal to let u equal the square root. The 1/sqrt(.) in the denominator is part of the derivative of the square root, which is exactly what you need for du.

Worked Example 9: Power of sec Times tan

Problem

\int \sec^7(x) \tan(x) \, dx

Identify the substitution

Rewrite as sec^6(x) · sec(x) tan(x) dx. Since the derivative of sec(x) is sec(x) tan(x), let u = sec(x).

Execute

• du = sec(x) tan(x) dx

• sec^7(x) tan(x) dx = sec^6(x) · sec(x) tan(x) dx = u^6 du

• ∫ u^6 du = u^7/7 + C

Result

\frac{\sec^7(x)}{7} + C

Why this works: The pattern generalises: ∫ sec^n(x) tan(x) dx = sec^n(x)/n + C for any n, because you always peel off one sec(x) tan(x) as du and the rest is a power of u. This same trick works for csc^n(x) cot(x) with u = csc(x).

Worked Example 10: Definite Integral and Odd Function Symmetry

Problem A: basic definite integral

\int_0^{\sqrt{\pi}} x \cos(x^2) \, dx

• Let u = x^2, du = 2x dx, so x dx = (1/2) du

• Change limits: x = 0 gives u = 0; x = sqrt(pi) gives u = pi

• = (1/2) ∫ from 0 to pi of cos(u) du = (1/2) [sin(u)] from 0 to pi

• = (1/2)(sin(pi) - sin(0)) = (1/2)(0 - 0) = 0

Result: 0

Problem B: odd function symmetry

\int_{-\pi/2}^{\pi/2} \frac{x^2 \sin(x)}{1 + x^8} \, dx

Identify the symmetry

Check each factor: x^2 is even, sin(x) is odd, 1/(1 + x^8) is even (since x^8 is even). The product of even · odd · even = odd. An odd function integrated over a symmetric interval [-a, a] gives zero.

Result: 0, with no computation needed.

Why this matters: On a timed exam, recognising symmetry can save several minutes of difficult algebra. Always check before you start substituting, especially when the limits are symmetric about zero.

Practice Q&A

Q: Evaluate ∫ cos(3x) / (1 + sin^2(3x)) dx.

A: Let u = sin(3x), du = 3cos(3x) dx. The integral becomes (1/3) ∫ du/(1 + u^2) = (1/3) arctan(sin(3x)) + C.

Q: Evaluate ∫ x^3 / (1 + x^8) dx.

A: Write x^8 = (x^4)^2. Let u = x^4, du = 4x^3 dx. The integral becomes (1/4) ∫ du/(1 + u^2) = (1/4) arctan(x^4) + C.

Q: Evaluate ∫ from 0 to 2 of x · e^(x^2) dx.

A: Let u = x^2, du = 2x dx. Limits become u = 0 to u = 4. The integral is (1/2) ∫ from 0 to 4 of e^u du = (1/2)(e^4 - 1).

Q: Is the function f(x) = x^5 / (1 + x^4) odd, even, or neither?

A: f(-x) = (-x)^5 / (1 + (-x)^4) = -x^5 / (1 + x^4) = -f(x). It is odd. So ∫ from -a to a of this function is zero.

Q: Evaluate ∫ sec^5(x) tan(x) dx.

A: Let u = sec(x), du = sec(x) tan(x) dx. The integral becomes ∫ u^4 du = u^5/5 + C = sec^5(x)/5 + C.

Q: Evaluate ∫ from 1 to e of (ln x)^2 / x dx.

A: Let u = ln x, du = (1/x) dx. Limits: x = 1 gives u = 0; x = e gives u = 1. The integral is ∫ from 0 to 1 of u^2 du = [u^3/3] from 0 to 1 = 1/3.

Q: What substitution would you use for ∫ sin(sqrt(x)) / sqrt(x) dx, and what is the result?

A: Let u = sqrt(x), du = 1/(2 sqrt(x)) dx, so dx/sqrt(x) = 2 du. The integral becomes 2 ∫ sin(u) du = -2 cos(sqrt(x)) + C.