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Trigonometric Substitution
In finding the area of a circle or an ellipse, an integral of the form arises,
where. If it were , the substitution would be effective
but, as it stands, is more difficult. If we change the variable from to by
the substitution , then the identity allows us to get rid of the
root sign because
Notice the difference between the substitution (in which the new variable is
a function of the old one) and the substitution (the old variable is a function of
the new one).
In general we can make a substitution of the form by using the Substitution
Rule in reverse. To make our calculations simpler, we assume that has an inverse func-
tion; that is, is one-to-one. In this case, if we replace by and by in the Substitution
Rule (Equation 5.5.4), we obtain
This kind of substitution is called inverse substitution.
We can make the inverse substitution provided that it defines a one-to-one
function. This can be accomplished by restricting to lie in the interval.
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 restric-
tion 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
EXAMPLE 1 Evaluate.
SOLUTION Let , where. Then and
(Note that because .) Thus, the Inverse Substitution Rule
gives
cot C
(^) y csc 2
1 d
(^) y
cos
2
sin
2
d (^) y cot 2
d
y
s 9 x^2
x
2 dx^ ^ y^
3 cos
9 sin
2
3 cos d
cos 0 2 2
s 9 x^2 s 9 9 sin 2 s9 cos 2 (^3) cos (^) 3 cos
x 3 sin 2 2 dx 3 cos d
y
s 9 x^2
x
2 dx
x a sin
y f^ x ^ dx^ ^ y f^ t t t t ^ dt
t u x x t
t
x t t
x a sin
u a
2
x
2
s a^2 x^2 s a^2 a^2 sin 2 s a^2 1 sin 2 s a^2 cos 2 a (^) cos (^)
1 sin
2
cos
2
x a sin
x s a^2 x^2 dx x
u a
2
x
2 a 0 x x s a^2 x^2 dx
x s a^2 x^2 dx
1
Expression Substitution Identity
x a sec , 0 sec 2 1 tan 2
or
s x^^2 ^ a^^2
x a tan , 1 tan 2 sec 2
s a^2 x^2
1 sin 2 cos 2 x a sin ,
s a^2 x^2
Since this is an indefinite integral, we must return to the original variable. This can be
done either by using trigonometric identities to express in terms of or
by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle.
Since , we label the opposite side and the hypotenuse as having lengths and.
Then the Pythagorean Theorem gives the length of the adjacent side as , so we
can simply read the value of from the figure:
(Although in the diagram, this expression for is valid even when .)
Since , we have and so
EXAMPLE 2 Find the area enclosed by the ellipse
SOLUTION Solving the equation of the ellipse for , we get
Because the ellipse is symmetric with respect to both axes, the total area 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
and so
To evaluate this integral we substitute. Then. To change the
limits of integration we note that when , , so ; when ,
, so. Also
since. Therefore
We have shown that the area of an ellipse with semiaxes and is. In particular,
taking , we have proved the famous formula that the area of a circle with
radius is.
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.
r
2
r
a b r
a b ab
ab
2 ab [
1
2 sin 2] 0
2
2 ab
4 ab y
2
0
cos
2
d 4 ab y
2
0
1
2 ^1 ^ cos 2^ d
A 4
b
a
y
a
0
s a^2 x^2 dx 4
b
a
y
2
0
a cos a cos d
s a^2 x^2 s a^2 a^2 sin 2 s a^2 cos 2 a cos a cos
sin 1 2
x 0 sin 0 0 x a
x a sin dx a cos d
1
4 A^ ^ y
a
0
b
a
s a^2 x^2 dx
y 0 x a
b
a
s a^2 x^2
A
y
b
a
or s a^2 x^2
y
2
b
2 ^1 ^
x
2
a
2 ^
a
2
x
2
a
2
y
x
2
a
2 ^
y
2
b
2 ^1
y
s 9 x^2
x^2
dx
s 9 x^2
x
sin^1
x
^ C
sin
1
sin x 3 x 3
0 cot 0
cot
s 9 x^2
x
cot
s 9 x^2
sin x 3 x 3
cot sin x 3
x
2 ■ TRIGONOMETRIC SUBSTITUTION
x
œ„„„„9-≈ „
FIGURE 1
sin ¨=
x 3
FIGURE 2
a@
b@
y
0 x
(0, b)
(a, 0)
4 ■ TRIGONOMETRIC SUBSTITUTION
The triangle in Figure 4 gives , so we have
Writing , we have
EXAMPLE 6 Find.
SOLUTION First we note that so trigonometric substitution
is appropriate. Although 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. When we combine this with the tangent substitution, we have ,
which gives and
When , , so ; when , , so.
Now we substitute so that. When , ; when
Therefore
EXAMPLE 7 Evaluate.
SOLUTION We can transform the integrand into a function for which trigonometric substi-
tution is appropriate by first completing the square under the root sign:
4 x 1
2
3 2 x x
2
3 x
2
2 x 3 1 x
2
2 x 1
y
x
s^3 ^2 x^ ^ x
2
dx
3
16 u^ ^
u
1
1 2
3 16 [(^
1 2 ^ 2)^ ^ ^1 ^1 ]^ ^
3 32
y
(^3) s 3 2
0
x
3
4 x
2
3 2 dx^ ^ ^
3
16 y
1 2
1
1 u
2
u
2 du^ ^
3
16 y
1 2
1
1 u
2
du
3, u
1 2
u cos du sin d 0 u 1
3
16 y
3
0
1 cos
2
cos
2
sin d
3
16 y
3
0
tan
3
sec
d
3
16 y
3
0
sin
3
cos
2
d
y
(^3) s 3 2
0
x
3
4 x
2
3 2 dx^ ^ y
3
0
27
8 tan^
3
27 sec
3
3
2 sec^
2
d
x 0 tan 0 0 x 3 s 3 2 tan s 3 3
s^4 x
s9 tan^
2 9 3 sec
dx
3
2 sec^
2
d
x
3
u 2 x 2 tan
s^4 x
4 x
2
3 2
s 4 x^2 9)
3
y
(^3) s 3 2
0
x
3
4 x
2
3 2 dx
y
dx
s x^2 a^2
ln x s x^2 a^2 C 1
C 1 C ln a
ln x s x^2 a^2 ln a C
y
dx
s x^2 a^2
ln
x
a
s x^2 a^2
a
^ C
tan s x^2 a^2 a
FIGU RE 4
sec ¨=
x a
œ „„„„„
a
x ≈-a@
This suggests that we make the substitution. Then and , so
y
x
s^3 ^2 x^ ^ x
2
dx y
u 1
s^4 ^ u
2
du
u x 1 du dx x u 1
TRIGONOMETRIC SUBSTITUTION ■ 5
We now substitute , giving and , so
s 3 2 x x^2 sin^1
x 1
^ C
s 4 u^2 sin^1
u
^ C
2 cos C
y 2 sin 1 d
y
x
s 3 2 x x^2
dx y
2 sin 1
2 cos
2 cos d
u 2 sin du 2 cos d s 4 u^2 2 cos
Exercises
1–3 Evaluate the integral using the indicated trigonometric
substitution. Sketch and label the associated right triangle.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
4–30 Evaluate the integral.
17. 18. (^) y
dx ax ^2 b^2 ^3 ^2 y
x
s x^^2 ^7
dx
y
dx
x 2 s 16 x^2 9
y
x 2
a 2 x 2 3 2 dx
y
du
u s 5 u^2
y
s x^^2 ^9 x 3 dx
y
1 0 y s^1 ^4 x^^2 dx^ x^ s x^^2 ^4 dx
y
t^5
s t^2 2
y dt
dx
s x^2 16
y
s x^2 a^2 x^4 y dx
x^2 s 25 x^2
dx
y
2 0 y^ x^^3 s x^^2 ^4 dx
2 s 2
t 3 s t^2 1
dt
y
(^2) s 3 0
x^3
s 16 x^2
dx
y x^ ^ 3 tan^
x 3
s x^^2 ^9
dx
y x^ x^ ^ 3 sin^
3 s 9 x^2 dx
y x^ ^ 3 sec^
x^2 s x^2 9
dx
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
31. (a) Use trigonometric substitution to show that
(b) Use the hyperbolic substitution to show that
These formulas are connected by Formula 3.9.3.
32. Evaluate
(a) by trigonometric substitution. (b) by the hyperbolic substitution x a sinh t.
y
x 2
x 2 a 2 3 2 dx
y
dx
s x^^2 ^ a^^2
sinh^1
x a
^ C
x a sinh t
y
dx
s x^2 a^2
ln( x (^) s x^2 a^2 ) C
y
2 0
cos t s 1 sin 2 t
y x^ s^1 ^ x^^4 dx dt
y
dx
5 4 x x 2
y (^5) 2
dx
x 2 2 x 2 2
y
x^2
s 4 x x^2
y dx
s 9 x^2 6 x 8
dx
y
dt
s t^^2 ^6 t^ ^13
y s^5 ^4 x^ ^ x^
(^2) dx
y
1 0 y s x^^2 ^1 dx
2 3 0
x^3 s 4 9 x^2 dx
y
t s 25 t^2
y dt
s 1 x^2 x
A Click here for answers. S Click here for solutions. dx
TRIGONOMETRIC SUBSTITUTION ■ 7
Answers
15. ( x s a^2 x^2 ) sin^1 x a C 17. s x^2 7 C
1 6 sec
(^1) x 3 s x (^2) 9 2 x (^2) C
1 4 sin
1 2 x
1 2 x^ s^1 ^4 x^ ln(s x^2 16 x ) C^2 C
24 s 3 8 s 25 x^2 25 x C
1 4
1 3 x^
2 s x^2 9 9 x C 18 s x^2 9 C
39. r (^) s R^2 r^2 r^2 2 R^2 arcsin r R 41. 2 2 Rr^2
1 6 (s^48 ^ sec
1
1 4 sin
(^1) x (^2) 1 4 x^
2 s^1 ^ x^^4 ^ C
1 2 tan
1 x 1 x 1 x 2 2 x 2 C
1 3 ln^ 3 x^ ^1 ^ s^9 x^ (^2) 6 x 8 ^ C
9 2 sin
1 x 2 3
1 2 ^ x^ ^2 s^5 ^4 x^ ^ x^
2 C
64 ln (^) (s 1 x (^) 1215 (^2) 1) x ^ s^1 ^ x^
2 C
S Click here for solutions.
1. Let x = 3 sec θ, where 0 ≤ θ < π 2 or^ π^ ≤^ θ^ <^
3 π
- Then dx = 3 sec θ tan θ dθ and p x^2 − 9 =
p 9 sec^2 θ − 9 =
p 9(sec^2 θ − 1) =
9 tan^2 θ
= 3 |tan θ| = 3 tan θ for the relevant values of θ.
Z
x^2
x^2 − 9
dx =
Z
9 sec^2 θ · 3 tan θ
3 sec θ tan θ dθ = 1 9
R
cos θ dθ = 1 9 sin^ θ^ +^ C^ =
x^2 − 9 x
+ C
Note that − sec(θ + π) = sec θ, so the figure is sufficient for the case π ≤ θ < 3 π
3. Let x = 3 tan θ, where − π 2 < θ < π 2. Then dx = 3 sec 2 θ dθ and
p x^2 + 9 =
p 9 tan^2 θ + 9 =
q 9(tan^2 θ + 1) =
9 sec^2 θ
= 3 |sec θ| = 3 sec θ for the relevant values of θ.
Z
x 3 √ x^2 + 9
dx =
Z
3 tan 3 θ 3 sec θ
3 sec 2 θ dθ = 3 3
Z
tan 3 θ sec θdθ = 3 3
Z
tan 2 θ tan θ sec θ dθ
3 R ¡
sec 2 θ − 1
tan θ sec θ dθ = 3
3 R ¡
u 2 − 1
du [u = sec θ, du = sec θ tan θ dθ]
3 ¡^1
3 u
3 − u
+ C = 3
3 ¡^1
3 sec
3 θ − sec θ
+ C = 3
3
x 2
x^2 + 9 3
+ C
1 3
x
2
p x^2 + 9 + C or 1 3
x
2 − 18
¢ p x^2 + 9 + C
5. Let t = sec θ, so dt = sec θ tan θ dθ, t =
2 ⇒ θ = π 4 , and^ t^ = 2^ ⇒^ θ^ =^
π
- Then
Z (^2)
√ 2
t^3
t^2 − 1
dt =
Z (^) π/ 3
π/ 4
sec^3 θ tan θ
sec θ tan θ dθ =
Z (^) π/ 3
π/ 4
sec^2 θ
dθ =
Z (^) π/ 3
π/ 4
cos 2 θ dθ
R (^) π/ 3 π/ 4
1 2 (1 + cos 2θ)^ dθ^ =^
1 2
θ + 1 2 sin 2θ
¤π/ 3 π/ 4
1 2
h³ π 3 +^
1 2
√ 3 2
¡ (^) π 4 +^
1 2 ·^1
¢i
1 2
π 12 +^
√ 3 4 −^
1 2
π 24 +^
√ 3 8 −^
1 4
7. Let x = 5 sin θ, so dx = 5 cos θ dθ. Then Z 1
x^2
25 − x^2
dx =
Z
52 sin^2 θ · 5 cos θ
5 cos θ dθ
1 25
R
csc 2 θ dθ = − 1 25 cot^ θ^ +^ C
25 − x^2 x
+ C
8 ■ TRIGONOMETRIC SUBSTITUTION
Solutions: Trigonometric Substitution
19. Let x = tan θ, where − π 2 <^ θ^ <^
π
- Then^ dx^ = sec
2 θ dθ
and
1 + x^2 = sec θ, so Z √ 1 + x^2 x
dx =
Z
sec θ tan θ
sec 2 θ dθ =
Z
sec θ tan θ
(1 + tan 2 θ) dθ
R
(csc θ + sec θ tan θ) dθ
= ln
1 + x^2 x
x
1 + x^2 1
1 + x^2 − 1 x
1 + x^2 + C
21. Let u = 4 − 9 x 2 ⇒ du = − 18 x dx. Then x 2 = 1 9 (4^ −^ u)^ and
R (^2) / 3 0 x
4 − 9 x^2 dx =
R 0
4
1 9 (4^ −^ u)u
1 18
du = 1 162
R 4
0
4 u 1 / 2 − u 3 / 2
du
1 162
h 8 3 u
3 / 2 − 2 5 u
5 / 2
i 4
0
1 162
64 3 −^
64 5
64 1215
Or: Let 3 x = 2 sin θ, where − π 2 ≤^ θ^ ≤^
π
23. 5 + 4x − x 2 = −(x 2 − 4 x + 4) + 9 = −(x − 2) 2 + 9. Let
x − 2 = 3 sin θ, − π 2 ≤ θ ≤ π 2 , so dx = 3 cos θ dθ. Then R √ 5 + 4x − x^2 dx =
R p 9 − (x − 2)^2 dx =
R p 9 − 9 sin 2 θ 3 cos θ dθ
=
R √
9 cos^2 θ 3 cos θ dθ =
R
9 cos 2 θ dθ
= 9 2
R
(1 + cos 2θ) dθ = 9 2
θ + 1 2 sin 2θ
+ C
9 2 θ^ +^
9 4 sin 2θ^ +^ C^ =^
9 2 θ^ +^
9 4 (2 sin^ θ^ cos^ θ) +^ C
sin − 1
μ x − 2 3
x − 2 3
5 + 4x − x^2 3
+ C
sin − 1
μ x − 2 3
(x − 2)
5 + 4x − x^2 + C
25. 9 x^2 + 6x − 8 = (3x + 1) 2 − 9 , so let u = 3x + 1, du = 3dx. Then
Z
dx √ 9 x^2 + 6x − 8
Z 1
√^3 du u^2 − 9
. Now
let u = 3 sec θ, where 0 ≤ θ < π 2 or^ π^ ≤^ θ^ <^
3 π
- Then^ du^ = 3 sec^ θ^ tan^ θ^ dθ^ and^
u^2 − 9 = 3 tan θ, so
Z (^1) 3 du √ u^2 − 9
Z
sec θ tan θ dθ 3 tan θ
1 3
R
sec θdθ = 1 3 ln|sec^ θ^ + tan^ θ|^ +^ C^1 =^
1 3 ln
u +
u^2 − 9 3
¯ +^ C^1
3 ln
¯u^ +^
p u^2 − 9
¯ +^ C^ =^
1 3 ln
¯^3 x^ + 1 +^
p 9 x^2 + 6x − 8
¯ +^ C
27. x 2 + 2x + 2 = (x + 1) 2 + 1. Let u = x + 1, du = dx. Then
Z dx
(x^2 + 2x + 2)
Z
du
(u^2 + 1)
Z
sec 2 θdθ sec^4 θ
where u = tan θ, du = sec 2 θ dθ, and u^2 + 1 = sec^2 θ
R
cos 2 θ dθ = (^12)
R
(1 + cos 2θ) dθ = 12 (θ + sin θ cos θ) + C
tan − 1 u +
u 1 + u^2
+ C =
tan − 1 (x + 1) +
x + 1 x^2 + 2x + 2
+ C
10 ■ TRIGONOMETRIC SUBSTITUTION
= ln |csc θ − cot θ| + sec θ + C [by Exercise 39 in Additional Topics: Trigonometric Integrals ]
29. Let u = x 2 , du = 2x dx. Then
R x
1 − x^4 dx =
R √
1 − u^2
1 2 du
1 2
R
cos θ · cos θ dθ
where u = sin θ, du = cos θ dθ,
and
1 − u^2 = cos θ
R
1 2 (1 + cos 2θ)dθ^ =^
1 4 θ^ +^
1 8 sin 2θ^ +^ C^ =^
1 4 θ^ +^
1 4 sin^ θ^ cos^ θ^ +^ C
= 1 4 sin
− 1 u + 1 4 u
1 − u^2 + C = 1 4 sin
− 1 (x 2 ) + 1 4 x
2
1 − x^4 + C
31. (a) Let x = a tan θ, where − π 2 <^ θ^ <^
π
- Then^
x^2 + a^2 = a sec θ and
Z
dx √ x^2 + a^2
Z
a sec 2 θ dθ a sec θ
Z
sec θ dθ = ln|sec θ + tan θ| + C 1 = ln
x^2 + a^2 a
x a
¯ +^ C^1
= ln
x +
p x^2 + a^2
(b) Let x = a sinh t, so that dx = a cosh t dt and
x^2 + a^2 = a cosh t. Then Z dx √ x^2 + a^2
Z
a cosh t dt a cosh t
= t + C = sinh
− 1 x a
+ C.
33. The average value of f(x) =
x^2 − 1 /x on the interval [1, 7] is
1 7 − 1
Z 7
1
x^2 − 1 x
dx =
Z (^) α
0
tan θ sec θ
· sec θ tan θ dθ
where x = sec θ, dx = sec θ tan θ dθ, √ x^2 − 1 = tan θ, and α = sec − 1 7
1 6
R (^) α 0 tan
2 θ dθ = 1 6
R (^) α 0 (sec
2 θ − 1) dθ
h tan θ − θ
iα
0
= 16 (tan α − α)
1 6
48 − sec − 1 7
35. Area of 4 P OQ = 12 (r cos θ)(r sin θ) = 12 r 2 sin θ cos θ. Area of region P QR =
R (^) r r cos θ
r^2 − x^2 dx.
Let x = r cos u ⇒ dx = −r sin u du for θ ≤ u ≤ π
- Then we obtain
R √
r^2 − x^2 dx =
R
r sin u (−r sin u) du = −r^2
R
sin^2 u du = − 12 r^2 (u − sin u cos u) + C
1 2 r
2 cos − 1 (x/r) + 1 2 x^
p r^2 − x^2 + C
so
area of region P QR = 1 2
h −r
2 cos
− 1 (x/r) + x
p r^2 − x^2
ir
r cos θ
= (^12)
−r 2 θ + r cos θ r sin θ
= 12 r 2 θ − 12 r 2 sin θ cos θ
and thus, (area of sector P OR) = (area of 4 P OQ) + (area of region P QR) = 1 2 r
2 θ.
TRIGONOMETRIC SUBSTITUTION ■ 11
