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Methods Of Integration Practice Test

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Review the following concepts if needed:

Methods Of Integration Practice Test

  1. Use integration by parts to evaluate each of the following indefinite integrals.

    (a) ∫ log2 x dx

    (b) ∫ x · e3x dx

    (c) ∫ x 2 cos x dx

    (d) ∫ t sin 3 t cos 3 t dt

    (e) ∫ cos y ln(sin y ) dy

    (f) ∫ x2 e4 x dx

  2. Use partial fractions to evaluate each of the following indefinite integrals.

    Methods of Integration 7.4 Integrals of Trigonometric Expressions

  3. Use the method of u -substitution to evaluate each of the following indefinite integrals.

    Methods of Integration 7.4 Integrals of Trigonometric Expressions

  4. Evaluate each of the following indefinite trigonometric integrals.

    (a) ∫ sin x cos2 x dx

    (b) ∫ sin3 x cos2 x dx

    (c) ∫ tan3 x sec2 x dx

    (d) ∫ tan x sec3 x dx

    (e) ∫ sin2 x cos2 x dx

    (f) ∫ sin x cos4 x dx

  5. Calculate each of the following definite integrals.

    Methods of Integration 7.4 Integrals of Trigonometric Expressions

Answers

  1. We do (a), (b), (c), (d).

    SOLUTIONS TO EXERCISES Chapter 7

    SOLUTIONS TO EXERCISES Chapter 7

     

    (d) Notice that SOLUTIONS TO EXERCISES Chapter 7 . Now let u = t and dv = sin 6 t d t . Then

    SOLUTIONS TO EXERCISES Chapter 7

  2. We do (a), (b), (c), (d).

    SOLUTIONS TO EXERCISES Chapter 7

    SOLUTIONS TO EXERCISES Chapter 7

  3. We do (a), (b), (c), (d).

    (a) Let u = sin x, du = cos x dx . Then the integral becomes

    SOLUTIONS TO EXERCISES Chapter 7

    Resubstituting x , we obtain the final answer

    SOLUTIONS TO EXERCISES Chapter 7

    (b) Let SOLUTIONS TO EXERCISES Chapter 7. Then the integral becomes

    ∫ 2 sin u du = −2 cos u + C.

    Resubstituting x , we obtain the final answer

    SOLUTIONS TO EXERCISES Chapter 7

    (c) Let u = ln x, du = [1/ x ] dx . Then the integral becomes

    SOLUTIONS TO EXERCISES Chapter 7

    Resubstituting x , we obtain the final answer

    SOLUTIONS TO EXERCISES Chapter 7

    (d) Let u = tan x, du = sec 2 x dx. Then the integral becomes

    e u du = eu + C.

    Resubstituting x, we obtain the final answer

    e tan x sec 2 x dx = e tan x + C.

  4. We do (a), (b), (c), (d).

    (a) Let u = cos x, du = − sin x dx. Then the integral becomes

    SOLUTIONS TO EXERCISES Chapter 7

    Resubstituting x , we obtain the final answer

    SOLUTIONS TO EXERCISES Chapter 7

    (b) Write

    SOLUTIONS TO EXERCISES Chapter 7

    Let u = cos x, du = − sin x dx . Then the integral becomes

    SOLUTIONS TO EXERCISES Chapter 7

    Resubstituting x , we obtain the final answer

    SOLUTIONS TO EXERCISES Chapter 7

    (c) Let u = tan x, du = sec 2 x dx . Then the integral becomes

    SOLUTIONS TO EXERCISES Chapter 7

    Resubstituting x , we obtain the final answer

    SOLUTIONS TO EXERCISES Chapter 7

    (d) Let u = sec x, du = sec x tan x . Then the integral becomes

    SOLUTIONS TO EXERCISES Chapter 7

    Resubstituting x , we obtain the final answer

    SOLUTIONS TO EXERCISES Chapter 7

  5. We do (a), (b), (c), (d).

    (a) Use integration by parts twice:

    SOLUTIONS TO EXERCISES Chapter 7

     

    We may now solve for the desired integral:

    SOLUTIONS TO EXERCISES Chapter 7

    (b) Integrate by parts with u = ln x, dv = x 2 dx . Thus

    SOLUTIONS TO EXERCISES Chapter 7

    (c) We write

    SOLUTIONS TO EXERCISES Chapter 7

    Thus

    SOLUTIONS TO EXERCISES Chapter 7

    (d) We write

    SOLUTIONS TO EXERCISES Chapter 7

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