Trigonometric Substitution - Stewart Calculus

Trigonometric Substitution

In finding the area of a circle or an ellipse, an integral of the form x sa 2 x 2 dx arises, where a 0. If it were x x sa 2 x 2 dx, the substitution u a 2 x 2 would be effective but, as it stands, x sa 2 x 2 dx is more difficult. If we change the variable from x to by

the substitution x a sin , then the identity 1 sin2 cos2 allows us to get rid of the root sign because

sa 2 x 2 sa 2 a 2 sin2 sa 21 sin 2 sa 2 cos2 a cos

Notice the difference between the substitution u a 2 x 2 (in which the new variable is a function of the old one) and the substitution x a sin (the old variable is a function of the new one).

In general we can make a substitution of the form x tt by using the Substitution Rule in reverse. To make our calculations simpler, we assume that t has an inverse function; that is, t is one-to-one. In this case, if we replace u by x and x by t in the Substitution Rule (Equation 5.5.4), we obtain

y f x dx y f tttt dt

This kind of substitution is called inverse substitution. We can make the inverse substitution x a sin provided that it defines a one-to-one

function. This can be accomplished by restricting to lie in the interval 2, 2. In the following table we list trigonometric substitutions that are effective for the given

radical expressions because of the specified trigonometric identities. In each case the restriction on is imposed to ensure that the function that defines the substitution is one-to-one. (These are the same intervals used in Appendix D in defining the inverse functions.)

Table of Trigonometric Substitutions

Expression sa 2 x 2 sa 2 x 2 sx 2 a 2

Substitution

x a sin ,

2

2

x a tan ,

2

2

x a sec , 0 or 3

2

2

Identity 1 sin2 cos2 1 tan2 sec2 sec2 1 tan2

s9 x 2

y EXAMPLE 1 Evaluate

x 2 dx.

SOLUTION Let x 3 sin , where 2 2. Then dx 3 cos d and

s9 x2 s9 9 sin2 s9 cos2 3 cos 3 cos

(Note that cos 0 because 2 2.) Thus, the Inverse Substitution Rule

gives

s9

y y x2

x2

dx

3 cos 9 sin2

3 cos

d

y y

cos2 sin2

d

cot2 d

y csc2 1 d

cot C

1

2 TRIGONOMETRIC SUBSTITUTION

3

? oe,,9,,-,,,,,,

FIGURE 1

sin

?=

x 3

y (0, b)

0

FIGURE 2

a@

+

? b@

=1

x

(a, 0) x

Since this is an indefinite integral, we must return to the original variable x. This can be done either by using trigonometric identities to express cot in terms of sin x3 or by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle. Since sin x3, we label the opposite side and the hypotenuse as having lengths x and 3. Then the Pythagorean Theorem gives the length of the adjacent side as s9 x 2, so we can simply read the value of cot from the figure:

cot s9 x 2 x

(Although 0 in the diagram, this expression for cot is valid even when 0.) Since sin x3, we have sin1x3 and so

y s9 x2 dx s9 x2 sin1 x C

x2

x

3

EXAMPLE 2 Find the area enclosed by the ellipse

x2

y2

a2 b2 1

SOLUTION Solving the equation of the ellipse for y, we get

y2 b2

1

x2 a2

a2 a2

x2

or

y b sa 2 x 2 a

Because the ellipse is symmetric with respect to both axes, the total area A is four times

the area in the first quadrant (see Figure 2). The part of the ellipse in the first quadrant is

given by the function

b y sa 2 x 2

a

0xa

and so

y 1

4

A

a b sa 2 x 2 dx 0 a

To evaluate this integral we substitute x a sin . Then dx a cos d. To change the limits of integration we note that when x 0, sin 0, so 0; when x a, sin 1, so 2. Also

sa 2 x 2 sa 2 a 2 sin2 sa 2 cos2 a cos a cos

since 0 2. Therefore

y y A 4 b a sa 2 x 2 dx 4 b 2 a cos a cos d

a 0

a 0

y y 4ab

2 cos2 d 4ab

2

1 2

1

cos

2

d

0

0

[ ]

2ab

1 2

sin

2

2 0

2ab

00 2

ab

We have shown that the area of an ellipse with semiaxes a and b is ab. In particular, taking a b r, we have proved the famous formula that the area of a circle with radius r is r 2.

NOTE Since the integral in Example 2 was a definite integral, we changed the limits of integration and did not have to convert back to the original variable x.

oe,,,,+,,,,4,,

? 2

FIGURE 3

tan

?=

x 2

TRIGONOMETRIC SUBSTITUTION 3

1

y EXAMPLE 3 Find

x

2

s

x

2

4

dx.

SOLUTION Let x 2 tan , 2 2. Then dx 2 sec2 d and

sx2 4 s4tan2 1 s4 sec2 2 sec 2 sec

Thus, we have

dx

y y y x2sx2

4

2 sec2 d 4 tan2 2 sec

1 4

sec tan2

d

To evaluate this trigonometric integral we put everything in terms of sin and cos :

sec

1 cos2 cos

tan2 cos sin2 sin2

Therefore, making the substitution u sin , we have

dx

1

y y y x

2

s

x

2

4

4

cos sin2

d

1 4

du u2

1 4

1 u

C

4

1 sin

C

x

csc C

4

We use Figure 3 to determine that csc sx 2 4x and so

y dx

x

2

s

x

2

4

sx2 4x

4

C

x

y EXAMPLE 4 Find

dx.

sx2 4

SOLUTION It would be possible to use the trigonometric substitution x 2 tan here (as in Example 3). But the direct substitution u x 2 4 is simpler, because then du 2x dx

and

x

1 du

y sx2 4 dx 2 y su su C sx2 4 C

NOTE Example 4 illustrates the fact that even when trigonometric substitutions are possible, they may not give the easiest solution. You should look for a simpler method first.

dx

y EXAMPLE 5 Evaluate

, where a 0. sx2 a2

SOLUTION We let x a sec , where 0 2 or 32. Then dx a sec tan d and

sx 2 a 2 sa 2sec 2 1 sa 2 tan2 a tan a tan

Therefore

y

dx sx2

a2

y

a

sec tan a tan

d

y sec d ln sec tan C

4 TRIGONOMETRIC SUBSTITUTION

x

? a

FIGURE 4

sec

?

=

x a

oe,,,,-,,,,a,,@

The triangle in Figure 4 gives tan sx 2 a 2a, so we have

y dx sx 2 a 2 ln

x sx2 a2

a

a

C

ln x sx2 a2 ln a C

Writing C1 C ln a, we have

y dx ln sx 2 a 2

x sx 2 a 2

C1

y EXAMPLE 6

Find

3 s32 0

4x 2

x3 932

dx.

SOLUTION First we note that 4x 2 932 s4x 2 9 )3 so trigonometric substitution

is appropriate. Although s4x 2 9 is not quite one of the expressions in the table of

trigonometric substitutions, it becomes one of them if we make the preliminary substitu-

tion

u

2x.

When

we

combine

this

with

the

tangent

substitution,

we

have

x

3 2

tan

,

which

gives

dx

3 2

sec2

d

and

s4x 2 9 s9 tan2 9 3 sec

When x 0, tan 0, so 0; when x 3s32, tan s3, so 3.

3 s32

y y 0

4x 2

x3 932

dx

3 0

27 8

tan3

27 sec3

3 2

sec2

d

y y

3 16

3 0

tan3 sec

d

3 16

3 0

sin3 cos2

d

y

3 16

3 0

1

cos2 cos2

sin

d

Now we substitute u cos so that du sin d. When 0, u 1; when 3, u 12.

Therefore

3 s32

y y y 0

4x 2

x3 932

dx

3 16

12 1

1

u2 u2

du

3 16

12 1 u 2 du

1

[( ) ]

3 16

u 1 u

12

3 16

1

1 2

2

1 1

3 32

x

y EXAMPLE 7 Evaluate

dx.

s3 2x x 2

SOLUTION We can transform the integrand into a function for which trigonometric substitution is appropriate by first completing the square under the root sign:

3 2x x 2 3 x 2 2x 3 1 x 2 2x 1 4 x 12

This suggests that we make the substitution u x 1. Then du dx and x u 1, so

x

u1

y s3 2x x2 dx y s4 u2 du

TRIGONOMETRIC SUBSTITUTION 5

We now substitute u 2 sin , giving du 2 cos d and s4 u 2 2 cos , so

y

s3

x 2x

x2

dx

y

2

sin 2 cos

1

2 cos d

y 2 sin 1 d

2 cos C

u

s4 u 2 sin1

C

2

s3 2x x2 sin1 x 1 C 2

Exercises

A Click here for answers.

S Click here for solutions.

1?3 Evaluate the integral using the indicated trigonometric substitution. Sketch and label the associated right triangle.

y 1.

1 x 2 sx 2 9 dx;

x 3 sec

y 2. x 3 s9 x 2 dx; x 3 sin

y 3.

x 3 dx; x 3 tan

sx 2 9

4?30 Evaluate the integral.

y 4.

2 s3

x3

dx

0 s16 x 2

y 5.

2

1

s2 t 3 st 2 1 dt

1

y 7. x 2 s25 x 2 dx

dx

y 9.

sx 2 16

y 6. 2 x 3 sx 2 4 dx 0

sx2 a2

y 8.

dx

x4

t5

y 10.

dt

st2 2

y 11. s1 4x 2 dx

sx 2 9

y 13.

x3 dx

y 12. 1 x sx 2 4 dx 0 du

y 14.

u s5 u 2

x2

y 15. a 2 x 2 32 dx

dx

y 16. x 2 s16x 2 9

x

y 17.

dx

sx 2 7

dx

y 18. ax2 b 2 32

s1 x 2

19. y x dx

t

y 20.

dt

s25 t 2

y 21. 23 x 3s4 9x 2 dx 0

y 22. 1 sx 2 1 dx 0

y 23. s5 4x x 2 dx

1

y 25.

dx

s9x 2 6x 8

dt

y 24.

st 2 6t 13

x2

y 26.

dx

s4x x 2

dx

y 27. x 2 2x 22

dx

y 28. 5 4x x 2 52

y 29. x s1 x 4 dx

2 cos t

y 30.

dt

0 s1 sin2t

31. (a) Use trigonometric substitution to show that

y dx

sx2

a2

ln(x

sx2

a2)

C

(b) Use the hyperbolic substitution x a sinh t to show that

y dx sinh1 x C

sx2 a2

a

These formulas are connected by Formula 3.9.3.

32. Evaluate

x2

y x 2 a 2 32 dx

(a) by trigonometric substitution. (b) by the hyperbolic substitution x a sinh t.

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