You will be able to
- Evaluate integrals using u-substitution, integration by parts, and trigonometric substitution
- Decompose rational functions with partial fractions and integrate the result
- Determine whether an improper integral converges or diverges and find its value when it converges
- Apply the ratio, root, comparison, and alternating series tests to decide convergence of a series
- Find Taylor and Maclaurin series and determine the interval of convergence of a power series
- Compute areas, volumes, and arc lengths with definite integrals
Good to know first
- Differentiation rules, including the chain and product rules
- Basic integration and the Fundamental Theorem of Calculus from Calculus 1
- Algebra skills for factoring and solving equations
- Trigonometric identities and values of common angles
Syllabus
9 lessons
- 1
Integration by parts: a first worked example (∫x² sin x dx) Free
This lesson shows how to use the integration by parts formula on a product of a polynomial and a trig function, the first problem of the Calculus 2 final review. You will see how to choose u and dv, find v, and set up the next step.
1 exercise3 questions - 2
Integration by Substitution (and Integration by Parts in Action) Beyond Pass
This lesson shows how to simplify a hard integral by swapping the variable x for a new variable u, so the integral becomes one you know how to solve. It also shows the parts-style trick used on x squared times sine x, and works a full substitution on cosine to the fifth times sine to the fourth.
2 code3 exercises3 questions - 3
Trigonometric Substitution and Powers of Sine and Cosine Beyond Pass
This lesson shows how to integrate a product of powers of sine and cosine by substituting u = sin x, and how to handle a radical like the square root of x squared plus 9 by substituting x = 3 tan θ. You will learn to choose the substitution that makes the integral simpler and to finish with the right u or θ and back-substitution.
1 code3 exercises4 questions - 4
Trigonometric substitution: integrating x³ over √(x²+9) Beyond Pass
You learn how to replace x with 3 tan θ so that a square root of x² + 9 turns into a simple secant, then how to draw a right triangle to turn the answer back into x. The worked example integrates x³ divided by √(x²+9) step by step.
3 formulas2 exercises3 questions - 5
Partial Fraction Decomposition Beyond Pass
Partial fraction decomposition splits a messy rational fraction into simpler fractions that can be integrated one by one. The video shows how to find the unknown constants by plugging in clever values of x.
2 exercises3 questions - 6
Partial fractions: integrating a rational function (the chapter's "improper integral" example) Beyond Pass
This lesson shows how to integrate a rational function by splitting it into simpler partial fractions and finding the unknown constants by plugging in clever x-values. The chapter's problem is an indefinite integral of a fraction, and the same decomposition method is what lets you integrate it term by term.
2 exercises3 questions - 7
Improper integrals, trapezoidal rule and Simpson's rule (review chapter) Beyond Pass
This lesson shows how to decide whether an improper integral converges by taking a limit of a proper integral, and how to estimate an integral from a table of values with the trapezoidal rule and Simpson's rule. Each method is applied to a worked example from the video.
1 code5 build steps3 exercises4 questions - 8
Simpson's Rule and Arc Length: Two Exam Problems Beyond Pass
This lesson walks through two final-exam problems: estimating displacement from a velocity table with Simpson's rule, and finding an arc length by substitution. Along the way it shows how units and the arc length formula guide each step.
2 exercises4 questions - 9
Surface area of revolution: rotating y = x^3 about the x-axis Beyond Pass
This lesson shows how to set up and evaluate the surface area of a solid made by spinning a curve around the x-axis, using the formula S = ∫ 2πf(x)√(1+[f'(x)]²) dx. The worked example rotates f(x) = x³ from x = 0 to x = 2, using a u-substitution to finish the integral.
1 code2 formulas3 exercises3 questions
Free preview · Lesson 1 in full
Integration by parts: a first worked example (∫x² sin x dx)
This lesson shows how to use the integration by parts formula on a product of a polynomial and a trig function, the first problem of the Calculus 2 final review. You will see how to choose u and dv, find v, and set up the next step.
Draw and write next to the video.
The idea
The video opens a Calculus 2 final exam review with problem 1: find the indefinite integral of x² sin x dx. The integrand is a product of two different kinds of functions, a power x² and a trig function sin x. Products like this cannot be integrated directly with the basic rules, so we use integration by parts.
The rule is written on the board as . The idea is to split the integrand into a part called u, which gets simpler when you differentiate it, and a part called dv, which you can integrate easily. Here the video picks u = x², so du = 2x dx, and dv = sin x dx, so v = −cos x.
Now substitute into the formula. The first term is u·v = x²(−cos x). The integral that remains is −∫ v du = −∫ (−cos x)(2x) dx. The power of x has dropped from 2 to 1, so the new integral is simpler than the old one.
Simplify the signs. Two negatives multiply to a positive, so the remaining integral becomes +∫ 2x cos x dx. The result so far is −x² cos x + ∫ 2x cos x dx. The video says this integral still needs integration by parts again, which is why the exam problem is a multi-step one.
A useful habit: choose u as the part that becomes simpler when differentiated (here x², which goes to 2x and then to 2). Choose dv as the part you can integrate (sin x integrates to −cos x). This choice is what makes the process shorten each time.
Key points
- Integration by parts formula: ∫u dv = uv − ∫v du. 1:00
- Pick u = x² (its derivative 2x is simpler) and dv = sin x dx. 1:00
- v is the antiderivative of dv: the antiderivative of sin x is −cos x. 1:15
- Substitute: x²(−cos x) − ∫(−cos x)(2x) dx. 1:41
- The two negatives cancel, leaving −x² cos x + ∫2x cos x dx. 2:20
Practice
Use integration by parts on ∫x sin x dx from 0 to π/2 with u = x and dv = sin x dx. Find the value of the definite integral. Recall that v = −cos x.
Hint
After the formula, the leftover integral is ∫cos x dx, which is sin x.
Solution
- Set u = x, so du = dx; dv = sin x dx, so v = −cos x.
- Apply the formula: ∫u dv = uv − ∫v du gives [−x cos x] + ∫cos x dx.
- Antiderivative: −x cos x + sin x, evaluated from 0 to π/2.
- At π/2: −(π/2)(0) + 1 = 1. At 0: 0 + 0 = 0.
- Difference: 1 − 0 = 1.
1
On your scratchpad
Write ∫u dv = uv − ∫v du in the corner of your scratchpad, then label u, du, dv and v for ∫x² sin x dx.
Quiz
In ∫x² sin x dx, why does the video choose u = x²?
- Because sin x is harder to differentiate
- Because x² becomes simpler when differentiated (to 2x)
- Because x² is always the dv part
- Because the answer must contain x²
Show the answer
B. Because x² becomes simpler when differentiated (to 2x) · Differentiating x² gives 2x, which is simpler, so the next integral is easier. Choosing u this way lowers the power of x.
1:00 watch the momentWhat is v when dv = sin x dx?
- v = cos x
- v = −cos x
- v = sin x
- v = x sin x
Show the answer
B. v = −cos x · The antiderivative of sin x is −cos x, so v = −cos x.
1:15 watch the momentAfter substituting into the formula, what is the remaining integral?
- ∫ x² cos x dx
- −∫(−cos x)(2x) dx
- ∫ 2x sin x dx
- ∫ sin x dx
Show the answer
B. −∫(−cos x)(2x) dx · The formula gives uv − ∫v du, with v = −cos x and du = 2x dx, so the remaining term is −∫(−cos x)(2x) dx.
1:41 watch the moment
Lessons 2–9
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Made by Beyond from the video “Calculus 2 Final Exam Review -” by The Organic Chemistry Tutor. The video plays in YouTube's own player. Watch it on YouTube · All lessons