Quiz 3 Quiz 3                              Friday, March 30, 2001

NAME:

Student Number:------            No.----           Marks ----

(25 Marks) Evaluate the following indefinite integral.

ó
õ
x2
 Ö 4-x2
dx

Solution. (Method 1) Let x = 2sint, p/2 £ t £ p/2. Then dx = 2cost dt and

x2
 Ö 4-x2
 =
4sin2t
 Ö 4-4sin2t
= 4sin2t
 2 Ö 1-sin2t
 =
2sin2t
 Ö cos2t
= 2sin2t
cost
.

Hence, we have

ó
õ
x2
 Ö 4-x2
dx
 =
 óõ 2sin2tcost .2cost dt = 4 óõ sin2t dt = 2 óõ (1-cos2t) dt
 =
 2(t- 12 sin2t)+C = 2t-sin2t +C = 2t-2sintcost+C

Since x = 2sint, p/2 £ t £ p/2, we have sint = x/2, t = arcsinx/2 and

 cost = Ö 1-sin2t = Ö 1-(x/2)2 = 12 Ö 4-x2 .

Hence, we obtain

ó
õ
x2
 Ö 4-x2
dx
 =
 2t-2sintcost+C
 =
 2arcsinx/2-2. x2 . 12 Ö 4-x2 +C
 =
 2arcsinx/2- x2 Ö 4-x2 +C.

(Mehtod 2). Use the known formulas:

ó
õ

Ö

a2-x2

dx = a2
2
arcsin x
a
+ x
2

Ö

a2-x2

and ó
õ
dx
 Ö a2-x2
= arcsin x
a
+C.

x2
 Ö 4-x2
 =
x2-4+4
 Ö 4-x2
= - 4-x2
 Ö 4-x2
+ 4
 Ö 4-x2
= -
Ö

4-x2

+4 1
 Ö 4-x2
 =
-
Ö

22-x2

+4 1
 Ö 22-x2
.

By the known formulas, we have

 óõ Ö 22-x2 dx = 222 arcsin x2 + x2 Ö 4-x2 +C1 = 2arcsin x2 + x2 Ö 4-x2 +C1

and

ó
õ
1
 Ö 22-x2
dx = arcsin x
2
+C2.

Hence, we have

ó
õ
x2
 Ö 4-x2
dx
 =
- ó
õ

Ö

22-x2

dx+4 ó
õ
1
 Ö 22-x2
dx
 =
 -2arcsin x2 - x2 Ö 4-x2 -C1+4arcsin x2 +4C2
 =
 2arcsin x2 - x2 Ö 4-x2 +C.

File translated from TEX by TTH, version 2.79.
On 3 Apr 2001, 10:34.