Part II: The Answers
476
Answers
701–800
2
1
2
2
1
0
2 4
0
2
2
0
1
1
0
1
1
1
x
dx
x
+
=
=
−
(
)
=
−
( )
=
∫
−
−
−
tan
tan
tan
π
π
764.
5
6
π
Here’s the given problem:
5
1
2
1 2
3 2
−
∫
x
dx
Because you have the antiderivative formula
1
1
2
1
−
=
+
−
∫
x
dx
x C
sin
, you can simply
apply that here to find the solution:
5
1
5
5
3
2
1
2
5 3 6
2
1 2
3 2
1
1 2
3 2
1
1
−
= (
)
=
−
=
−
∫
−
−
−
x
dx
x
sin
sin
s in
/
/
π π
( ( )
= 5
6
π
765.
1
3
3
1
sin
−
( )+
x C
You may want to write the second term underneath the radical as a quantity squared
so you can more easily see which substitution to use:
dx
x
dx
x
1 9
1 3
2
2
−
=
− ( )
∫
∫
Now begin by using the substitution u = 3x so that du = 3
dx, or 1
3
du dx
= . This gives
you the following:
dx
x
u
du
u C
x C
1 3
1
3
1
1
1
3
1
3
3
2
2
1
1
− ( )
=
−
=
( )+
=
( )+
∫
∫
−
−
sin
sin
476
Answers
701–800
2
1
2
2
1
0
2 4
0
2
2
0
1
1
0
1
1
1
x
dx
x
+
=
=
−
(
)
=
−
( )
=
∫
−
−
−
tan
tan
tan
π
π
764.
5
6
π
Here’s the given problem:
5
1
2
1 2
3 2
−
∫
x
dx
Because you have the antiderivative formula
1
1
2
1
−
=
+
−
∫
x
dx
x C
sin
, you can simply
apply that here to find the solution:
5
1
5
5
3
2
1
2
5 3 6
2
1 2
3 2
1
1 2
3 2
1
1
−
= (
)
=
−
=
−
∫
−
−
−
x
dx
x
sin
sin
s in
/
/
π π
( ( )
= 5
6
π
765.
1
3
3
1
sin
−
( )+
x C
You may want to write the second term underneath the radical as a quantity squared
so you can more easily see which substitution to use:
dx
x
dx
x
1 9
1 3
2
2
−
=
− ( )
∫
∫
Now begin by using the substitution u = 3x so that du = 3
dx, or 1
3
du dx
= . This gives
you the following:
dx
x
u
du
u C
x C
1 3
1
3
1
1
1
3
1
3
3
2
2
1
1
− ( )
=
−
=
( )+
=
( )+
∫
∫
−
−
sin
sin