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Calculus Volume 1

5.7 Integrals Resulting in Inverse Trigonometric Functions

Calculus Volume 15.7 Integrals Resulting in Inverse Trigonometric Functions
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 5.7.1. Integrate functions resulting in inverse trigonometric functions

In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before. Recall from Functions and Graphs that trigonometric functions are not one-to-one unless the domains are restricted. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also in Derivatives, we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions.

Integrals that Result in Inverse Sine Functions

Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.

Rule: Integration Formulas Resulting in Inverse Trigonometric Functions

The following integration formulas yield inverse trigonometric functions:


  1. dua2u2=sin−1u|a|+Cdua2u2=sin−1u|a|+C
    5.23

  2. dua2+u2=1atan−1ua+Cdua2+u2=1atan−1ua+C
    5.24

  3. duuu2a2=1|a|sec−1u|a|+Cduuu2a2=1|a|sec−1u|a|+C
    5.25

Proof

Let y=sin−1xa.y=sin−1xa. Then asiny=x.asiny=x. Now let’s use implicit differentiation. We obtain

ddx(asiny)=ddx(x)acosydydx=1dydx=1acosy.ddx(asiny)=ddx(x)acosydydx=1dydx=1acosy.

For π2yπ2,cosy0.π2yπ2,cosy0. Thus, applying the Pythagorean identity sin2y+cos2y=1,sin2y+cos2y=1, we have cosy=1sin2y.cosy=1sin2y. This gives

1acosy=1a1sin2y=1a2a2sin2y=1a2x2.1acosy=1a1sin2y=1a2a2sin2y=1a2x2.

Then for axa,axa, we have

1a2u2du=sin−1(ua)+C.1a2u2du=sin−1(ua)+C.

Example 5.49

Evaluating a Definite Integral Using Inverse Trigonometric Functions

Evaluate the definite integral 01dx1x2.01dx1x2.

Solution

We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have

01dx1x2=sin−1x|01=sin−11sin−10=π20=π2.01dx1x2=sin−1x|01=sin−11sin−10=π20=π2.

Checkpoint 5.40

Find the antiderivative of dx116x2.dx116x2.

Example 5.50

Finding an Antiderivative Involving an Inverse Trigonometric Function

Evaluate the integral dx49x2.dx49x2.

Solution

Substitute u=3x.u=3x. Then du=3dxdu=3dx and we have

dx49x2=13du4u2.dx49x2=13du4u2.

Applying the formula with a=2,a=2, we obtain

dx49x2=13du4u2=13sin−1(u2)+C=13sin−1(3x2)+C.dx49x2=13du4u2=13sin−1(u2)+C=13sin−1(3x2)+C.

Checkpoint 5.41

Find the indefinite integral using an inverse trigonometric function and substitution for dx9x2.dx9x2.

Example 5.51

Evaluating a Definite Integral

Evaluate the definite integral 03/2du1u2.03/2du1u2.

Solution

The format of the problem matches the inverse sine formula. Thus,

03/2du1u2=sin−1u|03/2=[sin−1(32)][sin−1(0)]=π3.03/2du1u2=sin−1u|03/2=[sin−1(32)][sin−1(0)]=π3.

Integrals Resulting in Other Inverse Trigonometric Functions

There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out −1 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.

Example 5.52

Finding an Antiderivative Involving the Inverse Tangent Function

Find an antiderivative of 11+4x2dx.11+4x2dx.

Solution

Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for tan−1u+C.tan−1u+C. So we use substitution, letting u=2x,u=2x, then du=2dxdu=2dx and 1/2du=dx.1/2du=dx. Then, we have

1211+u2du=12tan−1u+C=12tan−1(2x)+C.1211+u2du=12tan−1u+C=12tan−1(2x)+C.
Checkpoint 5.42

Use substitution to find the antiderivative of dx25+4x2.dx25+4x2.

Example 5.53

Applying the Integration Formulas

Find the antiderivative of 19+x2dx.19+x2dx.

Solution

Apply the formula with a=3.a=3. Then,

dx9+x2=13tan−1(x3)+C.dx9+x2=13tan−1(x3)+C.
Checkpoint 5.43

Find the antiderivative of dx16+x2.dx16+x2.

Example 5.54

Evaluating a Definite Integral

Evaluate the definite integral 3/33dx1+x2.3/33dx1+x2.

Solution

Use the formula for the inverse tangent. We have

3/33dx1+x2=tan−1x|3/33=[tan−1(3)][tan−1(33)]=π6.3/33dx1+x2=tan−1x|3/33=[tan−1(3)][tan−1(33)]=π6.
Checkpoint 5.44

Evaluate the definite integral 02dx4+x2.02dx4+x2.

Section 5.7 Exercises

In the following exercises, evaluate each integral in terms of an inverse trigonometric function.

391.

03/2dx1x203/2dx1x2

392.

−1/21/2dx1x2−1/21/2dx1x2

393.

31dx1+x231dx1+x2

394.

1/33dx1+x21/33dx1+x2

395.

12dx|x|x2112dx|x|x21

396.

12/3dx|x|x2112/3dx|x|x21

In the following exercises, find each indefinite integral, using appropriate substitutions.

397.

dx9x2dx9x2

398.

dx116x2dx116x2

399.

dx9+x2dx9+x2

400.

dx25+16x2dx25+16x2

401.

dx|x|x29dx|x|x29

402.

dx|x|4x216dx|x|4x216

403.

Explain the relationship cos−1t+C=dt1t2=sin−1t+C.cos−1t+C=dt1t2=sin−1t+C. Is it true, in general, that cos−1t=sin−1t?cos−1t=sin−1t?

404.

Explain the relationship sec−1t+C=dt|t|t21=csc−1t+C.sec−1t+C=dt|t|t21=csc−1t+C. Is it true, in general, that sec−1t=csc−1t?sec−1t=csc−1t?

405.

Explain what is wrong with the following integral: 12dt1t2.12dt1t2.

406.

Explain what is wrong with the following integral: −11dt|t|t21.−11dt|t|t21.

In the following exercises, solve for the antiderivative ff of f with C=0,C=0, then use a calculator to graph f and the antiderivative over the given interval [a,b].[a,b]. Identify a value of C such that adding C to the antiderivative recovers the definite integral F(x)=axf(t)dt.F(x)=axf(t)dt.

407.

[T] 19x2dx19x2dx over [−3,3][−3,3]

408.

[T] 99+x2dx99+x2dx over [−6,6][−6,6]

409.

[T] cosx4+sin2xdxcosx4+sin2xdx over [−6,6][−6,6]

410.

[T] ex1+e2xdxex1+e2xdx over [−6,6][−6,6]

In the following exercises, compute the antiderivative using appropriate substitutions.

411.

sin−1tdt1t2sin−1tdt1t2

412.

dtsin−1t1t2dtsin−1t1t2

413.

tan−1(2t)1+4t2dttan−1(2t)1+4t2dt

414.

ttan−1(t2)1+t4dtttan−1(t2)1+t4dt

415.

sec−1(t2)|t|t24dtsec−1(t2)|t|t24dt

416.

tsec−1(t2)t2t41dttsec−1(t2)t2t41dt

In the following exercises, use a calculator to graph the antiderivative ff with C=0C=0 over the given interval [a,b].[a,b]. Approximate a value of C, if possible, such that adding C to the antiderivative gives the same value as the definite integral F(x)=axf(t)dt.F(x)=axf(t)dt.

417.

[T] 1xx24dx1xx24dx over [2,6][2,6]

418.

[T] 1(2x+2)xdx1(2x+2)xdx over [0,6][0,6]

419.

[T] (sinx+xcosx)1+x2sin2xdx(sinx+xcosx)1+x2sin2xdx over [−6,6][−6,6]

420.

[T] 2e−2x1e−4xdx2e−2x1e−4xdx over [0,2][0,2]

421.

[T] 1x+xln2x1x+xln2x over [0,2][0,2]

422.

[T] sin−1x1x2sin−1x1x2 over [−1,1][−1,1]

In the following exercises, compute each integral using appropriate substitutions.

423.

ex1e2tdtex1e2tdt

424.

et1+e2tdtet1+e2tdt

425.

dtt1ln2tdtt1ln2t

426.

dtt(1+ln2t)dtt(1+ln2t)

427.

cos−1(2t)14t2dtcos−1(2t)14t2dt

428.

etcos−1(et)1e2tdtetcos−1(et)1e2tdt

In the following exercises, compute each definite integral.

429.

01/2tan(sin−1t)1t2dt01/2tan(sin−1t)1t2dt

430.

1/41/2tan(cos−1t)1t2dt1/41/2tan(cos−1t)1t2dt

431.

01/2sin(tan−1t)1+t2dt01/2sin(tan−1t)1+t2dt

432.

01/2cos(tan−1t)1+t2dt01/2cos(tan−1t)1+t2dt

433.

For A>0,A>0, compute I(A)=AAdt1+t2I(A)=AAdt1+t2 and evaluate limaI(A),limaI(A), the area under the graph of 11+t211+t2 on [,].[,].

434.

For 1<B<,1<B<, compute I(B)=1Bdttt21I(B)=1Bdttt21 and evaluate limBI(B),limBI(B), the area under the graph of 1tt211tt21 over [1,).[1,).

435.

Use the substitution u=2cotxu=2cotx and the identity 1+cot2x=csc2x1+cot2x=csc2x to evaluate dx1+cos2x.dx1+cos2x. (Hint: Multiply the top and bottom of the integrand by csc2x.)csc2x.)

436.

[T] Approximate the points at which the graphs of f(x)=2x21f(x)=2x21 and g(x)=(1+4x2)−3/2g(x)=(1+4x2)−3/2 intersect, and approximate the area between their graphs accurate to three decimal places.

437.

47. [T] Approximate the points at which the graphs of f(x)=x21f(x)=x21 and f(x)=x21f(x)=x21 intersect, and approximate the area between their graphs accurate to three decimal places.

438.

Use the following graph to prove that 0x1t2dt=12x1x2+12sin−1x.0x1t2dt=12x1x2+12sin−1x.


A diagram containing two shapes, a wedge from a circle shaded in blue on top of a triangle shaded in brown. The triangle’s hypotenuse is one of the radii edges of the wedge of the circle and is 1 unit long. There is a dotted red line forming a rectangle out of part of the wedge and the triangle, with the hypotenuse of the triangle as the diagonal of the rectangle. The curve of the circle is described by the equation sqrt(1-x^2).
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