techniques of integration

7TECHNIQUES OF INTEGRATIONTECHNIQUES OF INTEGRATION

TECHNIQUES OF INTEGRATION

7.3

Trigonometric Substitution

In this section, we will learn about:

The various types of trigonometric substitutions.

TRIGONOMETRIC SUBSTITUTION

In finding the area of a circle or an ellipse,

an integral of the form arises,

where a > 0.

If it were , the substitution would be effective.

However, as it stands, is more difficult.

2 2a x dx

2 2x a x dx 2 2 u a x

2 2 a x dx


If we change the variable from x to θ by

the substitution x = a sin θ, the identity

1 – sin2θ = cos2θ lets us lose the root sign.

This is because: 2 2 2 2 2

2 2

2 2

sin

(1 sin )

cos

cos

a x a a

a

a

a


Notice the difference between the substitution

u = a2 – x2 and the substitution x = a sin θ.

In the first, the new variable is a function of the old one.

In the second, the old variable is a function of the new one.


In general, we can make a substitution

of the form x = g(t)

by using the Substitution Rule in reverse.

To make our calculations simpler, we assume g has an inverse function, that is, g is one-to-one.

INVERSE SUBSTITUTION

Here, if we replace u by x and x by t in the

Substitution Rule (Equation 4 in Section 5.5),

we obtain:

This kind of substitution is called inverse substitution.

( ) ( ( )) '( )f x dx f g t g t dt

INVERSE SUBSTITUTION

We can make the inverse substitution

x = a sin θ, provided that it defines

a one-to-one function.

This can be accomplished by restricting θ to lie in the interval [-π/2, π/2].

TABLE OF TRIGONOMETRIC SUBSTITUTIONS

Here, we list trigonometric substitutions that

are effective for the given radical expressions

because of the specified trigonometric

identities.

TABLE OF TRIGONOMETRIC SUBSTITUTIONS

In each case, the restriction on θ is

imposed to ensure that the function that

defines the substitution is one-to-one. These are the same intervals used in Section 1.6

in defining the inverse functions.


Evaluate

Let x = 3 sin θ, where –π/2 ≤ θ ≤ π/2.

Then, dx = 3 cos θ dθ and

Note that cos θ ≥ 0 because –π/2 ≤ θ ≤ π/2.)

Example 1

2

2

9

xdx

x

2 2 29 9 9sin 9cos 3 cos 3cos x


Thus, the Inverse Substitution Rule

gives: 2

2 2

2

2

2

2

9 3cos3cos

9sin

cos

sin

cot

(csc 1)

cot

xdx d

x

d

d

d

C

Example 1


As this is an indefinite integral, we must

return to the original variable x.

This can be done in either of two ways.

Example 1


One, we can use trigonometric

identities to express cot θ in terms

of sin θ = x/3.

Example 1


Two, we can draw a diagram, where

θ is interpreted as an angle of a right

triangle.

Example 1


Since sin θ = x/3, we label the opposite

side and the hypotenuse as having lengths

x and 3.

Example 1


Then, the Pythagorean Theorem gives

the length of the adjacent side as:29 x

Example 1


So, we can simply read the value of cot θ

from the figure:

Although θ > 0 here, this expression for cot θ is valid even when θ < 0.

Example 1

29cot

x

x


As sin θ = x/3, we have θ = sin-1(x/3).

Hence,

2 21

2 2

9 9sin

3

x x xdx C

x x

Example 1


Find the area enclosed by

the ellipse 2 2

2 21

x y

a b

Example 2


Solving the equation of the ellipse for y,

we get

or

2 2 2 2

2 2 21

y x a x

b a a

Example 2

2 2by a x

a


As the ellipse is symmetric with respect

to both axes, the total area A is four times

the area in the first quadrant.

Example 2


The part of the ellipse in the first quadrant

is given by the function

Hence,

2 2 0 b

y a x x aa

2 214 0

a b

A a x dxa

Example 2


To evaluate this integral,

we substitute x = a sin θ.

Then, dx = a cos θ dθ.

Example 2


To change the limits of integration,

we note that:

When x = 0, sin θ = 0; so θ = 0

When x = a, sin θ = 1; so θ = π/2

Example 2


Also, since 0 ≤ θ ≤ π/2,

2 2 2 2 2

2 2

sin

cos

cos

cos

a x a a

a

a

a

Example 2


Therefore, 2 2

0

/ 2

0

/ 2 2

0

/ 2120

/ 2102

4

4 cos cos

4 cos

4 (1 cos 2 )

2 [ sin 2 ]

2 0 02

abA a x dx

ab

a a da

ab d

ab d

ab

ab ab

Example 2


We have shown that the area of an ellipse

with semiaxes a and b is πab.

In particular, taking a = b = r, we have proved the famous formula that the area of a circle with radius r is πr2.

Example 2


The integral in Example 2 was a definite

integral.

So, we changed the limits of integration,

and did not have to convert back to

the original variable x.

Note


Find

Let x = 2 tan θ, –π/2 < θ < π/2.

Then, dx = 2 sec2 θ dθ and

2 2

1

4 dxx x

2 2

2

4 4(tan 1)

4sec

2 sec

2sec

x

Example 3


Thus, we have:

2

22 2

2

2sec

4 tan 2sec41 sec

4 tan

dx d

x x

d

Example 3


To evaluate this trigonometric integral,

we put everything in terms of sin θ and

cos θ:

2

2 2 2

sec 1 cos cos

tan cos sin sin

Example 3


Therefore, making the substitution u = sin θ,

we have:22 2

2

1 cos

4 sin41

41 1

4

1 csc

4sin 4

dxd

x xdu

u

Cu

C C

Example 3zz


We use

the figure to

determine that:

Hence,

2csc 4 /x x

2

2 2

4

44

dx xC

xx x

Example 3


Find

It would be possible to use the trigonometric substitution x = 2 tan θ (as in Example 3).

2 4

xdx

x

Example 4


However, the direct substitution u = x2 + 4 is simpler.

This is because, then, du = 2x dx and

2

2

1

24

4

x dudx

ux

u C

x C

Example 4


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.

Note


Evaluate

where a > 0.

2 2

dx

x a

Example 5


We let x = a sec θ, where 0 < θ < π/2 or

π < θ < π/2.

Then, dx = a sec θ tan θ dθ and

E. g. 5—Solution 1

2 2 2 2

2 2

(sec 1)

tan

tan tan

x a a

a

a a


Therefore,

2 2

sec tan

tan

sec

ln sec tan

dx ad

ax a

d

C



The triangle in the figure gives:

2 2tan /x a a



So, we have:

2 2

2 2

2 2

ln

ln ln

dx x x aC

a ax a

x x a a C



Writing C1 = C – ln a, we have:

2 212 2

lndx

x x a Cx a

E. g. 5—Sol. 1 (For. 1)


For x > 0, the hyperbolic substitution

x = a cosh t can also be used.

Using the identity cosh2y – sinh2y = 1, we have:

2 2 2 2

2 2

(cosh 1)

sinh

sinh

x a a t

a t

a t



Since dx = a sinh t dt,

we obtain:

2 2

sinh

sinh

dx a t dt

a tx a

dt

t C



Since cosh t = x/a, we have t = cosh-1(x/a)

and

1

2 2cosh

dx xC

ax a

E. g. 5—Sol. 2 (For. 2)


Although Formulas 1 and 2 look quite

different, they are actually equivalent by

Formula 4 in Section 3.11

E. g. 5—Sol. 2 (For. 2)


As Example 5 illustrates, hyperbolic

substitutions can be used instead of

trigonometric substitutions, and sometimes

they lead to simpler answers.

However, we usually use trigonometric substitutions, because trigonometric identities are more familiar than hyperbolic identities.

Note


Find

First, we note that

So, trigonometric substitution is appropriate.

33 3 / 2

2 3/ 20 (4 9)

xdx

x

2 3/ 2 2 3(4 9) ( 4 9) x x

Example 6


is not quite one of the

expressions in the table of trigonometric

substitutions.

However, it becomes one if we make the preliminary substitution u = 2x.

24 9x Example 6


When we combine this with the tangent

substitution, we have .

This gives and

32 tanx

2 24 9 9 tan 9

3sec

x

Example 6

232 secdx d


When x = 0, tan θ = 0; so θ = 0.

When x = , tan θ = ; so θ = π/3.

Example 6

3 3 / 2 3

TRIGONOMETRIC SUBSTITUTION33 273 3 / 2 /3 28 3

22 3/ 2 30 0

3/3

316 0

3/3

316 20

2/3316 20

tansec

(4 9) 27sec

tan

sec

sin

cos

1 cossin

cos

xdx d

x

d

d

d

Example 6


Now, we substitute u = cos θ so that

du = - sin θ dθ.

When θ = 0, u = 1.

When θ = π/3, u = ½.

Example 6


Therefore,

33 3 / 2

2 3/ 20

21/ 2316 21

1/ 2 2316 1

1/ 2

316

1

3 3116 2 32

(4 9)

1

(1 )

1

2 (1 1)

xdx

x

udu

u

u du

uu

Example 6


Evaluate

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

23 2

xdx

x x

Example 7

2 2

2

2

3 2 3 ( 2 )

3 1 ( 2 1)

4 ( 1)

x x x x

x x

x


This suggests we make the substitution

u = x + 1.

Then, du = dx and x = u – 1.

So, 2 2

1

3 2 4

x udx du

x x u

Example 7


We now substitute .

This gives

and

2sinu

2cosdu d 24 2cosu

Example 7


So,2

2 1

2 1

2sin 12cos

2cos3 2

(2sin 1)

2cos

4 sin2

13 2 sin

2

xdx d

x x

d

C

uu C

xx x C

Example 7


The figure shows the graphs of the integrand

in Example 7 and its indefinite integral (with

C = 0).

Which is which?

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