CramX
RegisterLog in
Back to AI Flashcard MakerMathematics /AP Calculus AB: 9.3.1 Integrating Composite Trigonometric Functions by Substitution

AP Calculus AB: 9.3.1 Integrating Composite Trigonometric Functions by Substitution

Mathematics12 CardsCreated 3 months ago

This content covers the technique of integration by substitution applied to composite trigonometric functions. It explains how to identify the inner function for substitution, handle constant multiples, and ensure complete substitution. Worked examples demonstrate how substitution simplifies complex integrals, with emphasis on converting back to the original variable and verifying results through differentiation.

PrintEmbedImportReport

Integrating Composite Trigonometric Functions by Substitution

  • Integration by substitution is a technique for finding the antiderivative of a composite function. A composite function is a function that results from first applying one function, then another.

  • If the du-expression is only off by a constant multiple, you can still use integration by substitution by moving that constant out of the integral.

Tap or swipe ↕ to flip
Swipe ←→Navigate
1/12

Key Terms

Term
Definition

Integrating Composite Trigonometric Functions by Substitution

  • Integration by substitution is a technique for finding the antiderivative of a composite function. A composite function is a function that ...

note

  • This integral involves a composite function: the sine of a complicated expression. If you let u be the inside of the function, notice that ...

Solve the integral:∫x^−1/4 csc^2x^3/4dx

−4/3cotx^3/4+C

Integrate:∫3x^2sinx^3dx

− cos (x ^3 ) + C

Integrate.∫2t(1+t^2)^2sec^2[(1+t^2)^3]dt

1/3tan[(1+t^2)^3]+C

Find the integral.∫5xcosx^2dx

5/2sin  x^2+C

Log in to view all terms

Related Flashcard Decks

Mathematics

Essential SAT Math Formulas

This deck covers key formulas and concepts essential for the SAT Math section, including slope, distance, and vertex forms.

10 cards
View Deck
Mathematics

Algebra 1 Unit 1 Mastery Test

These properties of equality help maintain balance in an equation when performing the same operation on both sides. Inverse operations, like using addition to undo subtraction, are key tools in solving equations.

29 cards
View Deck
Mathematics

ATI TEAS Test Practice Mathematics Part 2

This flashcard set includes a variety of practical math questions covering percentages, conversions, averages, scale drawings, and basic algebra. It’s designed to strengthen problem-solving skills for real-life and standardized test situations.

53 cards
View Deck
Mathematics

ATI TEAS Test Practice Mathematics Part 1

This flashcard set includes a variety of practical math questions covering percentages, conversions, averages, scale drawings, and basic algebra. It’s designed to strengthen problem-solving skills for real-life and standardized test situations.

55 cards
View Deck
Mathematics

Unit 2 Progress Check: MCQ Part A

To determine the point where the tangent line to the graph of the function 𝑓f has a slope of 2 for 𝑥 >0 x>0, we use the fact that the slope of the tangent line at any point is given by the derivative 𝑓′(𝑥)f′(x).

15 cards
View Deck
Mathematics

Customary Liquid Measurement Units

This set of flashcards outlines key relationships between customary units of liquid measurement, including pints, cups, quarts, and gallons. It provides conversion facts to help understand how these units relate to one another in the U.S. measurement system.

3 cards
View Deck

Study Tips

  • Press F to enter focus mode for distraction-free studying
  • Review cards regularly to improve retention
  • Try to recall the answer before flipping the card
  • Share this deck with friends to study together
AP Calculus AB: 9.3.1 Integrating Composite Trigonometric Functions by Substitution
TermDefinition

Integrating Composite Trigonometric Functions by Substitution

  • Integration by substitution is a technique for finding the antiderivative of a composite function. A composite function is a function that results from first applying one function, then another.

  • If the du-expression is only off by a constant multiple, you can still use integration by substitution by moving that constant out of the integral.

note

  • This integral involves a composite function: the sine of a complicated expression. If you let u be the inside of the function, notice that du is found surrounding the sine function.

  • After you substitute u, make sure that nothing remains in terms of x.

  • Recall that the derivative of –cosx is sinx.

  • Make sure to replace u with its expression in terms of x.

  • You can check that your answer is correct by taking its
    derivative.

  • Here is another composite function. Let u be the inside
    expression. When you find du, you will notice that there is no multiple of 4 in the integrand, just dx.

  • Since 4 is just a constant multiple, solve for dx and substitute that expression into the integrand.

  • You can move 1/4 outside the integrand since it is a constant multiple.

  • After you integrate, make sure to replace u with its expression in terms of x.

  • Take the derivative of your answer to make sure it is correct.

Solve the integral:∫x^−1/4 csc^2x^3/4dx

−4/3cotx^3/4+C

Integrate:∫3x^2sinx^3dx

− cos (x ^3 ) + C

Integrate.∫2t(1+t^2)^2sec^2[(1+t^2)^3]dt

1/3tan[(1+t^2)^3]+C

Find the integral.∫5xcosx^2dx

5/2sin  x^2+C

Evaluate:∫2x^3sinx^4dx.

−1/2cosx^4+C


Evaluate the integral.

∫cos√x/√x dx

2sin√x+C

Integrate:∫xsec^2(x^2−1)dx.

1/2tan(x^2−1)+C

Evaluate:∫√x cscx^3/2 cotx^3/2dx

−2/3cscx^3/2+C

Evaluate the integral:∫^3√x⋅sec^2(1−x^4/3)dx


−3/4tan(1−x^4/3)+C

Solve the integral.∫(sec2xtan2x) dx

sec2x/2+C