Part II: The Answers
510
Answers
801–900
851.
2 2 2
−
The given integral is
dx
x
x
e
e
ln
2
∫
Begin with the u-substitution u = ln x so that du x
dx
= 1 . You can also compute the new
limits of integration by noting that if x = e, then u = ln e = 1, and if x = e
2
, then u = ln e
2
= 2.
Now you can integrate:
1
1 2
2
2 2 2 1
2 2 2
1
2
1 2
1
2
1 2
1
2
1
2
u
du
u du
u
u
∫
∫
=
=
=
=
−
=
−
−
852.
− (
) +
1
2
2
ln(cos )
θ
C
Here’s the given integral:
tan ln(cos )
θ
θ θ
d
∫
Start with the substitution u = ln(cos )
θ so that du
d
d
=
−
= −
1
cos
( sin )
t an
θ
θ θ
θ θ.
Substituting these values into the original integral, you get the following:
−
=−
+
= − (
) +
∫ udu u C
C
2
2
2
1
2
ln(cos )
θ
853.
−
+
−
1
3
3
e
C
x
The given integral is
x e dx
x
2
3
−
∫
Begin by letting u = –x
3
so that du = –3x
2 
dx, or −
=
1
3
2
du x dx. Substitute these values
into the original integral:
−
( ) = − +
= −
+
∫
−
1
3
1
3
1
3
3
e du
e C
e
C
u
u
x