Integration Requiring Multiple Techniques
Z
∞
0
∞
Z
0
Z
lim
a→∞
Z
0
a
x arctan x
dx
(1 + x2 )2
x arctan x
dx
(1 + x2 )2
x arctan x
dx
(1 + x2 )2
=
=
x =
dx =
tan θ.
sec2 θ dθ.
u =
du =
θ.
dθ.
b
θ tan θ
sec2 θ
2
2
(1
+
tan
θ)
0
Z b
θ tan θ
lim
sec2 θ
b→π/2 0 (sec2 θ)2
Z b
θ tan θ
lim
sec2 θ
b→π/2 0 sec4 θ
Z b
θ tan θ
lim
b→π/2 0 sec2 θ
Z b
θ sin θ cos2 θ
lim
cos θ
b→π/2 0
Z b
lim
θ sin θ cos θ
lim
dθ
=
dθ
=
dθ
=
dθ
=
dθ
=
dθ
=
b→π/2
b→π/2
0
b Z b
1
θ
2 sin θ −
sin2 θ dθ
lim
2
b→π/2
0 2
0
b
Z
θ
1 b1
lim
sin2 θ −
(1 − cos(2θ)) dθ
2 0 2
b→π/2 2
0
b
1
1
θ
sin2 θ −
θ − sin(2θ) dθ
lim
4
2
b→π/2 2
0
dv
v
=
=
sin θ cos θ dθ.
2
1
2 sin θ.
=
=
=
=
=
=
1
1
θ
sin2 b −
b − sin(2b) −
lim
4
2
b→π/2 2
0
1
1
sin2 0 −
0 − sin(0)
2
4
2
π π
−
4
8
π
.
8
For this approach we need the following.
Z
x
dx =
(1 + x2 )2
Z
1
2x
dx =
2
(1 + x2 )2
Z
1
u−2 du =
2
=
∞
Z
0
Z
lim
a→∞
0
a
u = 1 + x2 .
du = 2x dx.
1
− u−1 + C
2
1
−
+ C.
2(1 + x2 )
x arctan x
dx
(1 + x2 )2
=
x arctan x
dx
(1 + x2 )2
=
u =
du =
arctan x.
1
1+x2 dx.
=
x
dx
tan θ.
sec2 θ dθ.
a Z a
− arctan x 1
lim
−
−
dx
a→∞ 2(1 + x2 ) 2(1
+
x2 )2
0
0
a
Z
− arctan x 1 b
1
−
lim
lim
−
sec2 θ dθ
2
2
2
a→∞ 2(1 + x )
2
b→π/2
(1
+
tan
θ)
0
0
a
Z
− arctan x 1 b
1
−
lim
lim
−
sec2 θ dθ
2 θ)2
a→∞ 2(1 + x2 ) 2
(sec
b→π/2
0
0
a
Z
− arctan x 1 b
1
lim
−
lim
− 2 dθ
a→∞ 2(1 + x2 ) sec θ
b→π/2 2 0
0
a
Z b
1
− arctan x +
lim
cos2 θ dθ
lim
a→∞ 2(1 + x2 ) 2
b→π/2
0
0
a
Z b
− arctan x 1
1
lim
+
lim
(1 + cos(2θ)) dθ
a→∞ 2(1 + x2 ) 2 b→π/2 0 2
0
b
a
− arctan x 1
1
lim
+
lim
θ
+
sin(2θ)
a→∞ 2(1 + x2 )
4 b→π/2
2
0
0
− arctan a 0
lim
− +
a→∞ 2(1 + a2 )
2
1
1
lim b + sin(2b) − 0
4 b→π/2
2
=
=
=
=
=
=
=
=
=
=
π π
−
4
8
π
.
8
dv
v
x
= (1+x
2 )2 dx.
1
= − 2(1+x
2) .