Calculus II ep02: u-substitution (Sep 10, 2026)

Overview

Prof Staecker teaches u-substitution as reversing the chain rule: identify an inner expression, set it equal to u, convert its derivative into du, integrate in u, and substitute back. Worked examples progress from power and trigonometric integrals to definite integrals and harder cases requiring constant factors, solving for x, or rewriting extra powers; Prof Staecker also explains how to check an antiderivative by differentiation.

Key takeaways

A useful u-substitution typically pairs an inner expression with its derivative: setting u = x² + 10 makes 2x dx become du.
When the derivative of the chosen u differs by a constant, isolate that factor: for u = 1 − x³, x² dx = −du/3.
For definite integrals, either keep the original x-bounds until the antiderivative is rewritten in x, or convert the bounds to u-values; never apply x-bounds directly to a u-antiderivative.
If substitution leaves extra x factors, solve for x or an appropriate power of x in terms of u; for u = x² + 1, x² becomes u − 1.
Differentiate an indefinite-integral answer to check it; u-substitution problems commonly require the chain rule during verification.
Maintain the integral sign through substitution and algebraic rearrangement, and remove it only when the antiderivative is actually evaluated.

Chapters

0:00 Homework and the Wednesday Homework–Thursday Quiz Schedule

The first homework is posted through the class website, QR code, or Blackboard and is due Wednesday.
Thursday quizzes cover the material on the previous week's homework, so homework grades may not be returned before the quiz.
Prof Staecker introduces u-substitution as the second item on the course topic checklist.

1:18 U-Substitution Reverses the Chain Rule

The central idea is to change the variable inside an integral so a complicated expression becomes simpler.
For an expression such as (5x + 2)^6, setting u = 5x + 2 makes the outer power straightforward to integrate.
The chain rule contributes the derivative of the inside; an antiderivative needs that factor present or must account for it with a constant.

9:30 A Four-Step Workflow for Choosing and Using u

Choose u as the inner expression, often the contents of parentheses or a radical.
Differentiate the choice to obtain du, including dx in the differential notation.
Rewrite every x and dx in terms of u and du, integrate, then replace u with the original expression.
A poor choice of u may leave x-dependent pieces that cannot be rewritten cleanly.

11:00 Matching 2x dx in the Integral of (x² + 10)⁶

For ∫2x(x² + 10)^6 dx, choose u = x² + 10, giving du = 2x dx.
The substitution converts the integral directly to ∫u^6 du, with no extra constant adjustment.
Integrating and substituting back gives (x² + 10)^7/7 + C.

16:20 Constant Factors, Differentials, and a Square-Root Example

For ∫x²√(1 − x³) dx, choose u = 1 − x³, so du = −3x² dx and x² dx = −du/3.
Rewrite √u as u^(1/2); integration gives −(1/3)(2/3)u^(3/2).
The resulting antiderivative is −(2/9)(1 − x³)^(3/2) + C.
Keep integral signs during substitution steps; the differential disappears when the integral is evaluated.

24:48 Trig Substitution with cos x in the Denominator

For ∫sin x/cos^4 x dx, set u = cos x, so du = −sin x dx.
The numerator factor sin x dx matches −du, leaving −∫u^(−4) du.
Integrating and substituting back yields 1/(3 cos^3 x) + C.
The notation cos^4 x means (cos x)^4, not repeated composition of cosine.

30:10 Definite Integrals: Keep the Original x-Bounds Until the End

For ∫₁⁴ x(3x² + 5)^10 dx, use u = 3x² + 5 and x dx = du/6.
Integrating in u and substituting back gives (3x² + 5)^11/66.
Prof Staecker's method retains the original bounds 1 and 4 until the antiderivative is back in x.
The evaluated result is [(3x² + 5)^11/66] from x = 1 to x = 4; no +C is needed for a definite integral.

34:30 Independent Practice: Two U-Substitution Integrals

Students work on two practice problems, simplify their answers, and can compare approaches with classmates.
Prof Staecker circulates to offer help while students work rather than collecting or grading the exercises.
The examples practice recognizing an inner function and matching its derivative to the remaining factors.

41:21 Practice Solutions: Sine of x² and a Negative-Power Integral

For ∫₀^√π x sin(x²) dx, set u = x²; then x dx = du/2 and the evaluated integral is 1.
For ∫x²/(1 − x³)^7 dx, set u = 1 − x³, so x² dx = −du/3.
The second integral becomes −(1/3)∫u^(−7) du and yields 1/[18(1 − x³)^6] + C.
A sign error in a worked line is corrected during the first solution; the final value remains 1.

49:55 Check an Antiderivative by Differentiating It

Differentiate 1/[18(1 − x³)^6] using the chain rule to verify the result.
The derivative contributes −6(1 − x³)^(−7) and the inner derivative −3x².
The constants and signs simplify to x²/(1 − x³)^7, confirming the original integrand.

52:37 Harder Substitution: Rewrite x in ∫x√(x + 1) dx

Choose u = x + 1, giving du = dx, then solve for x = u − 1.
Rewrite the integral as ∫(u − 1)u^(1/2) du and distribute to obtain u^(3/2) − u^(1/2).
Integrating and substituting back gives (2/5)(x + 1)^(5/2) − (2/3)(x + 1)^(3/2) + C.
This example shows that u-substitution can require rewriting leftover x factors in terms of u.

58:00 Two Challenge Problems Require Rewriting Leftover x Terms

Students attempt two harder integrals, with the first modeled on the preceding square-root example.
Both problems require an extra algebraic step beyond matching the inner expression with du.
For the first challenge, Prof Staecker advises solving for x before completing the substitution.

1:05:15 Challenge Solution: Convert x + 5 Using u = x − 7

For ∫(x + 5)(x − 7)^3 dx, choose u = x − 7 and solve for x = u + 7.
The remaining factor becomes x + 5 = u + 12, so the integral is ∫(u + 12)u^3 du.
Integrating and substituting back gives (x − 7)^5/5 + 3(x − 7)^4 + C.

1:07:12 Challenge Solution: Separate x³ into x² and x

For ∫x³/(x² + 1)^10 dx, set u = x² + 1, giving x dx = du/2.
Split x³ dx into x²(x dx); replace x² with u − 1 and x dx with du/2.
The integral becomes (1/2)∫(u^10 − u^9) du, yielding (1/2)(u^11/11 − u^10/10) + C.
Substitute u = x² + 1 in the final answer; the homework is due Wednesday.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Prof Staecker.