Lecture 9Section 8.4 Integrals Involving √a2 − x2, √a2 + x2,
√
x2 − a2 Trigonometric Substitutions
Jiwen He
1
1.1
Trigonometric Substitution
Sine Substitution
Sine Substitution:
Sine Substitution:
√
√
1 − x2
1 − x2
1 − sin2 u = cos2 u
x = sin u
dx = cos u du
1 − x2 = cos2 u
1
Example 1.
Z √
Z
Z p
2
2
1 − x dx =
cos u cos u du = cos2 u du
Z 1 1
1
1
=
+ cos 2u du = u + sin 2u + C
2 2
2
4
1
= (u + sin u cos u) + C
2
p
1
sin−1 x + x 1 − x2 + C
=
2
Sine Substitution:
Sine Substitution:
√
√
a2 − x2
a2 − x2
1 − sin2 u = cos2 u
x = a sin u
dx = a cos u du
a2 − x2 = a2 cos2 u
2
Example 2.
Z p
Z √
Z
a2 cos2 u a cos u du = a2 cos2 u du
Z 1 1
a2
a2
2
=a
+ cos 2u du = u +
sin 2u + C
2 2
2
4
a2
(u + sin u cos u) + C
=
2
p
a2 −1
sin x + x 1 − x2 + C
=
2
a2 − x2 dx =
Example: Area of a Circle
Sine Substitution:
√
a2 − x2
1 − sin2 u = cos2 u
x = a sin u
dx = a cos u du
a2 − x2 = a2 cos2 u
Example 3.
Z r p
Z
r2 − x2 dx =
π
2
√
r2 cos2 u r cos u du = r2
−π
2
−r
=r
2
1
1
u + sin 2u
2
4
π
2
cos2 u du
−π
2
π2
=
−π
2
Z
πr2
2
= area enclosed by a semicircle of radius r
3
Example: Completing the Square
Sine Substitution:
p
a2 − (x − c)2
1 − sin2 u = cos2 u
x − c = a sin u
dx = a cos u du
a2 − (x − c)2 = a2 cos2 u
Example 4.
Z p
Z p
x2
2 − + 4x dx =
2 + 4 − (x2 + 4x + 4) dx
Z
Z p
6
=
6 − (x − 2)2 dx = 6 cos2 u du = (u + sin u cos u) + C
2
x − 2p
−1 x − 2
2
√ +
= 3 sin
6 − (x − 2) + C
6
6
√
√
x − 2 = 6 sin u, dx = 6 cos u du, 6 − (x − 2)2 = 6 cos2 u
Sine Substitution:
√
a2 − x2
4
Sine Substitution:
√
a2 − x2
1 − sin2 u = cos2 u
x = a sin u
dx = a cos u du
a2 − x2 = a2 cos2 u
Example 5.
Z √
1.2
Z √
Z
4 cos2 u
cos2 u
2 cos u du = 2
du
2 sin u
sin u
Z
Z
1 − sin2 u
du = 2 csc u − sin u du
=2
sin u
= 2 ln | csc u − cot u| + 2 cos u + C
√
p
2
4 − x2
| + 4 − x2 + C = · · ·
= 2 ln | −
x
x
4 − x2
dx =
x
Tangent Substitution
Tangent Substitution: 1 + x2
5
Tangent Substitution: 1 + x2
1 + tan2 u = sec2 u
x = tan u
dx = sec2 u du
1 + x2 = sec2 u
Example 6.
Z
Z
Z
1
2
sec u du = cos2 u du
4u
sec
Z 1 1
1
1
=
+ cos 2u du = u + sin 2u + C
2 2
2
4
1
= (u + sin u cos u) + C
2
1
1
x
=
tan−1 x + √
·√
+ C = ···
2
1 + x2
1 + x2
1
dx =
(1 + x2 )2
Tangent Substitution:
√
a2 + x2
6
Tangent Substitution:
√
a2 + x2
1 + tan2 u = sec2 u
x = a tan u
dx = a sec2 u du
2
a + x2 = a2 sec2 u
Example 7.
Z p
Z √
Z
a2 + x2 dx =
a2 sec2 ua sec2 u du = a2 sec3 u du
a2
(sec u tan u + ln | sec u + tan u|) + C
2 √
√
!
a2 + x2
2
x a
a2 + x2 x
+ ln + + C = ···
=
2
a
a
a
a
=
Example
7
Tangent Substitution:
√
a2 + x2
1 + tan2 u = sec2 u
x = a tan u
dx = a sec2 u du
a2 + x2 = a2 sec2 u
Example 8.
Z
1
1
√
dx =
2
2
x 9 + 4x
Z
1
q
3
sec2 u du
2
tan u
sec2 u
Z
Z
1
sec u
1
=
du =
csc u du
3
tan u
3
1
= ln | csc u − cot u| + C
3
√
1
9 + 4x2 − 3
= ln |
|+C
3
2x
3
2
Example
8
9
4
Tangent Substitution:
√
a2 + x2
1 + tan2 u = sec2 u
x = a tan u
dx = a sec2 u du
2
a + x2 = a2 sec2 u
Example 9.
Z
Z
Z
4 tan2 u
8 tan2 u sec2 u
x2
2
√
√
√
du
dx =
2
sec
u
du
=
2
4 + x2
4
+
4
tan
u
4 sec2 u
Z
Z
2
2
8 tan u sec u
=
du = 4 tan2 u sec u du
2 sec u
finish as in Example 12, Section 8.3
substitute in terms of x
Example: Completing the Square
9
Tangent Substitution:
p
a2 + (x − c)2
1 + tan2 u = sec2 u
x − c = a tan u
dx = a sec2 u du
2
a + (x − c)2 = a2 sec2 u
Example 10.
Z
Z p
p
2
x 16 + x − 6x dx = x 16 − 9 + (x2 − 6x + 9) dx
Z p
= x 7 + (x − 3)2 dx
Z √
√
√
7 tan u + 3
7 sec2 u 7 sec2 u du
=
Finish it!
1.3
Secant Substitution
Secant Substitution: x2 − 1
10
Secant Substitution: x2 − 1
sec2 u − 1 = tan2 u
x = sec u
dx = sec u tan u du
x2 − 1 = tan2 u
Example 11.
Z
Z
Z
sec u tan u
1
1
√
√
dx =
du =
du
2
2
2
2
sec
u
x x −1
Z sec u tan u
=
cos u du = sin u + C
√
=
Secant Substitution:
√
x2 − 1
+C
x
x2 − a2
11
Secant Substitution:
√
x2 − a2
sec2 u − 1 = tan2 u
x = a sec u
dx = a sec u tan u du
x2 − a2 = a2 tan2 u
Example 12.
Z
Z
2 sec u tan u
1
1
du
=
du
2
3/2
8
tan2 u
Z2 sec u(4 tan u) Z
1
1
=
cot2 u du =
csc2 u − 1 du
8
8
1
= (− cot u − u) + C
8 1
2
−1
√
=−
+ sec (x/2) + C
8
x2 − 4
1
dx =
2
x(x − 4)3/2
Z
Example: Completing the Square
12
Secant Substitution:
√
x2 − a2
sec2 u − 1 = tan2 u
x − c = a sec u
dx = a sec u tan u du
2
(x − c) − a2 = a2 tan2 u
Example 13.
Z
Z
x
x
√
p
dx =
dx
2 + 2x + 1) − 316 − 1
x2 + 2x − 3
(x
Z
x
p
, dx
=
(x + 1)2 − 4
Z
2 sec u − 1
√
2 sec u tan u du
=
4 tan2 u
Finish it!
Summary
a2 − x2
1 − sin2 u = cos2 u
x = a sin u
dx = a cos u du
a2 − x2 = a2 cos2 u
13
sin u =
x
a
√
a2 − x2
a
x
tan u = √
a2 − x2
cos u =
a2 + x2
1 + tan2 u = sec2 u
x = a tan u
dx = a sec2 u du
2
a + x2 = a2 sec2 u
14
x
a2 + x2
a
cos u = √
2
a + x2
x
tan u =
a
sin u = √
x2 − a2
sec2 u − 1 = tan2 u
x = a sec u
dx = a sec u tan u du
x2 − a2 = a2 tan2 u
√
sin u =
cos u =
a
x
√
tan u =
x2 − a2
x
x2 − a2
a
Outline
Contents
1 Trigonometric Substitution
1.1 Sine Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Tangent Substitution . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Secant Substitution . . . . . . . . . . . . . . . . . . . . . . . . . .
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1
5
10