Answers and Explanations 505
Answers
801–900
The second integral is
tan sec
θ
θ θ
π
π
2
3
3
d
−
∫
To evaluate the second integral, use the substitution u = tanθ so that du
d
= sec
2
θ θ.
Notice that you can find the new limits of integration by noting that if θ π
= 3
, then
tan π
3
3
( ) = , and if θ π
= − 3
, then tan −
( ) =
π
3
3. Therefore, the second integral
becomes
tan sec
θ
θ θ
π
π
2
3
3
3
3
0
d
u du
−
∫
∫
=
=
Combining the values of the first and second integrals gives you the answer:
tan
3
3
3
0
θ θ
π
π
−
∫
=
d
839.
4
The given integral is
(ln )
x
x
dx
e
3
1
2
∫
Start by using the substitution u = ln x so that du x
dx
= 1 . You can find the new limits of
integration by noting that if x = e
2
, then u = ln e
2
= 2, and if x = 1, then u = ln 1 = 0.
Therefore, the integral becomes
(ln )
x
x
dx
u du
u
e
3
1
3
0
2
4
0
2
4
2
4
2
4
0
4
∫
∫
=
=
=
−
=
840.
−
−
+
1
3
4 3
ln
x C
The given integral is
dx
x
4 3
−
∫
Start by letting u = 4 – 3x so that du = –3dx, or −
=
1
3
du dx. Substitute this back into the
original integral:
−
=−
+
= −
−
+
∫
1
3
1
1
3
1
3
4 3
u
du
u C
x C
ln
ln