Answers and Explanations 587
Answers
901–1,001
2
4 2
2
4 1
2
4
2
2
2
2
cos
sin
cos
sin
cos
cos
co
θ θ
θ
θ θ
θ
θ θ
θ
d
d
d
− (
)
=
−
(
)
=
(
)
=
∫
∫
∫
s s
cos
θ θ
θ
θ
θ
d
d
C
2
1
∫
∫
=
= +
From the substitution x = 2sinθ , or x
2
= sinθ , you have sin
−
( ) =
1
2
x
θ ; therefore,
θ + =
( ) +
−
C
x C
sin
1
2
Because
dx
x
x C
4
2
2
1
−
=
( ) +
∫
−
sin
, it follows that
lim
lim sin
a
a
a
a
dx
x
x
→
→
−
−
−
−
=
( )


 


 
∫
2
2
0
2
1
0
4
2
Evaluating the limit gives you
lim sin
lim sin
s in
s
a
a
a
x
a
→
−
→
−
−
−
−
( )


 


 
=
( ) −






=
2
1
0
2
1
1
2
2
0
i in
sin
−
−
( ) −
=
( )
=
1
1
2
2
0
1
2
π
The answer is finite, so the integral is convergent.