Quiz 3 Quiz 3                              Friday, March 30, 2001



NAME:

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






(25 Marks) Evaluate the following indefinite integral.



x2



4-x2
 dx

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


x2



4-x2
=
4sin2t



4-4sin2t
= 4sin2t
2


1-sin2t
=
2sin2t



cos2t
= 2sin2t
cost
.

Hence, we have



x2



4-x2
 dx
=

2sin2t
cost
.2cost dt = 4
sin2t dt = 2
(1-cos2t) dt
=
2(t- 1
2
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 =

 

1-sin2t
 
=

 

1-(x/2)2
 
= 1
2


 

4-x2
 
.

Hence, we obtain



x2



4-x2
 dx
=
2t-2sintcost+C
=
2arcsinx/2-2. x
2
. 1
2


 

4-x2
 
+C
=
2arcsinx/2- x
2


 

4-x2
 
+C.


(Mehtod 2). Use the known formulas:





 

a2-x2
 
 dx = a2
2
arcsin x
a
+ x
2


 

a2-x2
 
   and
dx



a2-x2
= arcsin x
a
+C.


x2



4-x2
=
x2-4+4



4-x2
= - 4-x2



4-x2
+ 4



4-x2
= -

 

4-x2
 
+4 1



4-x2
=
-

 

22-x2
 
+4 1



22-x2
.

By the known formulas, we have





 

22-x2
 
 dx = 22
2
arcsin x
2
+ x
2


 

4-x2
 
+C1 = 2arcsin x
2
+ x
2


 

4-x2
 
+C1

and



1



22-x2
 dx = arcsin x
2
+C2.

Hence, we have



x2



4-x2
 dx
=
-


 

22-x2
 
 dx+4
1



22-x2
 dx
=
-2arcsin x
2
- x
2


 

4-x2
 
-C1+4arcsin x
2
+4C2
=
2arcsin x
2
- x
2


 

4-x2
 
+C.



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