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Back to FlashcardsMathematics / 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 8 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.

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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.

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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

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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