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Calculus Volume 2

1.7 Integrals Resulting in Inverse Trigonometric Functions

Calculus Volume 21.7 Integrals Resulting in Inverse Trigonometric Functions

Learning Objectives

  • 1.7.1 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. Assume a>0a>0:


  1. dua2u2=1asin-1ua+Cdua2u2=1asin-1ua+C
    (1.23)

  2. dua2+u2=1atan−1ua+Cdua2+u2=1atan−1ua+C
    (1.24)

  3. duuu2a2=1|a|sec−1|u|a+Cduuu2a2=1|a|sec−1|u|a+C
    (1.25)

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, and generalizing to u, we have

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

Example 1.49

Evaluating a Definite Integral Using Inverse Trigonometric Functions

Evaluate the definite integral 012dx1x2.012dx1x2.

Checkpoint 1.40

Evaluate the integral dx116x2.dx116x2.

Example 1.50

Finding an Antiderivative Involving an Inverse Trigonometric Function

Evaluate the integral dx49x2.dx49x2.

Checkpoint 1.41

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

Example 1.51

Evaluating a Definite Integral

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

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 1.52

Finding an Antiderivative Involving the Inverse Tangent Function

Evaluate the integral 11+4x2dx.11+4x2dx.

Checkpoint 1.42

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

Example 1.53

Applying the Integration Formulas

Evaluate the integral 19+x2dx.19+x2dx.

Checkpoint 1.43

Evaluate the integral dx16+x2.dx16+x2.

Example 1.54

Evaluating a Definite Integral

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

Checkpoint 1.44

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

Section 1.7 Exercises

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

391.

0 3 / 2 d x 1 x 2 0 3 / 2 d x 1 x 2

392.

−1 / 2 1 / 2 d x 1 x 2 −1 / 2 1 / 2 d x 1 x 2

393.

3 1 d x 1 + x 2 3 1 d x 1 + x 2

394.

1 / 3 3 d x 1 + x 2 1 / 3 3 d x 1 + x 2

395.

2 3 2 d x | x | x 2 1 2 3 2 d x | x | x 2 1

396.

2 2 d x | x | x 2 1 2 2 d x | x | x 2 1

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

397.

d x 9 x 2 d x 9 x 2

398.

d x 1 16 x 2 d x 1 16 x 2

399.

d x 9 + x 2 d x 9 + x 2

400.

d x 25 + 16 x 2 d x 25 + 16 x 2

401.

d x | x | x 2 9 d x | x | x 2 9

402.

d x | x | 4 x 2 16 d x | x | 4 x 2 16

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 −1 t d t 1 t 2 sin −1 t d t 1 t 2

412.

d t sin −1 t 1 t 2 d t sin −1 t 1 t 2

413.

tan −1 ( 2 t ) 1 + 4 t 2 d t tan −1 ( 2 t ) 1 + 4 t 2 d t

414.

t tan −1 ( t 2 ) 1 + t 4 d t t tan −1 ( t 2 ) 1 + t 4 d t

415.

sec −1 ( t 2 ) | t | t 2 4 d t sec −1 ( t 2 ) | t | t 2 4 d t

416.

t sec −1 ( t 2 ) t 2 t 4 1 d t t sec −1 ( t 2 ) t 2 t 4 1 d t

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.

e t 1 e 2 t d t e t 1 e 2 t d t

424.

e t 1 + e 2 t d t e t 1 + e 2 t d t

425.

d t t 1 ln 2 t d t t 1 ln 2 t

426.

d t t ( 1 + ln 2 t ) d t t ( 1 + ln 2 t )

427.

cos −1 ( 2 t ) 1 4 t 2 d t cos −1 ( 2 t ) 1 4 t 2 d t

428.

e t cos −1 ( e t ) 1 e 2 t d t e t cos −1 ( e t ) 1 e 2 t d t

In the following exercises, compute each definite integral.

429.

0 1 / 2 tan ( sin −1 t ) 1 t 2 d t 0 1 / 2 tan ( sin −1 t ) 1 t 2 d t

430.

1 / 4 1 / 2 tan ( cos −1 t ) 1 t 2 d t 1 / 4 1 / 2 tan ( cos −1 t ) 1 t 2 d t

431.

0 1 / 2 sin ( tan −1 t ) 1 + t 2 d t 0 1 / 2 sin ( tan −1 t ) 1 + t 2 d t

432.

0 1 / 2 cos ( tan −1 t ) 1 + t 2 d t 0 1 / 2 cos ( tan −1 t ) 1 + t 2 d t

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 to four decimal places, and approximate the area between their graphs to three decimal places.

437.

47. [T] Approximate the points at which the graphs of f(x)=x21f(x)=x21 and g(x)=(x2+1)12g(x)=(x2+1)12 intersect to four decimal places, and approximate the area between their graphs 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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