THE COLLEGE FINALS
⇒
The Finals will be conducted in rounds. One at a time, each
remaining contestant will have two and a half minutes to
compute an indefinite integral. If answered correctly, the
contestant remains in the competition. Once every remaining
contestant has a empted one problem, a round is completed. If
during any round, all contestants are unable to complete a
problem correctly, all contestants will remain in the competition
for another round.
The last person remaining wins an additional $75 and will be
crowned the Integration Champion!
2 0 1 3
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I N T E G R A T I O N
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INTEGRAL #1
READY,
GET SET,…
2 : 30
2 0 1 3
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of
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I N T E G R A T I O N
B E E
INTEGRAL #1
∫
ln x
x2
dx
2 : 30
2 0 1 3
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of
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I N T E G R A T I O N
B E E
INTEGRAL #1
∫
ln x
x2
dx
[
integrate by parts:
=−
ln x
= −
2 0 1 3
x
∫
+
u = ln x
du =
1
x
dx
,
dv =
1
x2
dx
]
v = − x1
1
ln x 1
dx = −
− +C
x2
x
x
1 + ln x
+C
x
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #2
READY,
GET SET,…
2 : 30
2 0 1 3
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of
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I N T E G R A T I O N
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INTEGRAL #2
∫
(
x
e +e
)
−x 2
dx
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #2
∫
(
x
e +e
∫
=
(
)
−x 2
dx
−2x
2x
e +2+e
)
e2x
e−2x
=
+ 2x −
+C
2
2
4x + e2x − e−2x
=
+C
2
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #3
READY,
GET SET,…
2 : 30
2 0 1 3
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I N T E G R A T I O N
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INTEGRAL #3
∫
√
e2x 1 + e2x dx
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #3
∫
√
e2x 1 + e2x dx
∫
[
1 √
=
u du u = 1 + e2x,
2
2x
du = 2e dx
]
u3/2
=
+C
3
(
=
)
2x 3/2
1+e
3
2 0 1 3
U
+C
of
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I N T E G R A T I O N
B E E
INTEGRAL #4
READY,
GET SET,…
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #4
∫
√
ln x
x
dx
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #4
∫
√
ln x
x
∫
dx
∫
1 ln x
dx =
dx
=
x
2 x
[
]
∫
1
1
=
u du u = ln x, du = dx
2
x
ln x1/2
u2
(ln x)2
=
=
+C
4
4
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #5
READY,
GET SET,…
2 : 30
2 0 1 3
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I N T E G R A T I O N
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INTEGRAL #5
∫
e2x − e−2x
dx
2x
−2x
e +e
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #5
∫
e2x − e−2x
dx
2x
−2x
e +e
1
=
2
=
∫
[
1
du u = e2x + e−2x,
u
2x
du = 2e − 2e
−2x
)
]
dx
1
ln u + C
2
(
=
(
ln e
2 0 1 3
2x
−2x
+e
2
U
of
)
+C
S
I N T E G R A T I O N
B E E
INTEGRAL #6
READY,
GET SET,…
2 : 30
2 0 1 3
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of
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I N T E G R A T I O N
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INTEGRAL #6
∫
x sec2 x dx
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #6
∫
x sec2 x dx
[
integration by parts:
∫
u = x
du = dx
2
,
dv = sec x dx
]
v = tan x
= x tan x − tan x dx
= x tan x + ln|cos x| + C
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #7
READY,
GET SET,…
2 : 30
2 0 1 3
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I N T E G R A T I O N
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INTEGRAL #7
)3
∫ (
1
dx
x 1+
x
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #7
)3
∫ (
1
dx
x 1+
x
)
∫ (
3
3
1
= x 1 + + 2 + 3 dx
x x
x
)
∫(
3
1
=
x + 3 + + 2 dx
x x
x2
1
=
+ 3x + 3 ln x − + C
2
x
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #8
READY,
GET SET,…
2 : 30
2 0 1 3
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of
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I N T E G R A T I O N
B E E
INTEGRAL #8
∫
cos5 x dx
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #8
∫
cos5 x dx
∫
=
∫
=
∫
=
(
(
(
2 0 1 3
2
cos x
1−u
∫
)2
cos x dx =
)
2 2
(
)2
1 − sin x cos x dx
du [u = sin x,
2
du = cos x dx]
)
sin5 x 2 sin3 x
u − 2u + 1 du =
−
+ sin x + C
5
3
4
2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #9
READY,
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2 : 30
2 0 1 3
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I N T E G R A T I O N
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INTEGRAL #9
∫
x2 + 1
dx
2
(x + 1)
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #9
∫
x2 + 1
dx
2
(x + 1)
∫
(u − 1)2 + 1
=
du [u = x + 1, du = dx,
2
u
)
∫ 2
∫(
u − 2u + 2
2
2
=
du
=
1
−
+
du
2
2
u
u u
= u − 2 ln|u| −
2 0 1 3
U
of
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x = u − 1]
2
2
+ C = x + 1 − 2 ln|x + 1| −
+C
u
x+1
I N T E G R A T I O N
B E E
INTEGRAL #10
READY,
GET SET,…
2 : 30
2 0 1 3
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of
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I N T E G R A T I O N
B E E
INTEGRAL #10
∫
x2
dx
x−1
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #10
∫
x2
dx
x−1
∫(
x+1+
=
1
x−1
)
dx
x2
=
+ x + ln|x − 1| + C
2
2 0 1 3
U
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I N T E G R A T I O N
B E E
INTEGRAL #11
READY,
GET SET,…
2 : 30
2 0 1 3
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of
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I N T E G R A T I O N
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INTEGRAL #11
∫
1
√
dx
x2 − 1
2 : 30
2 0 1 3
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of
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I N T E G R A T I O N
B E E
INTEGRAL #11
∫
1
√
dx
x2 − 1
[
∫
=
x = sec θ,
dx = sec θ tan θ dθ,
sec θ tan θ
dθ =
tan θ
√
]
2
x − 1 = tan θ
∫
sec θ dθ
√
= ln|sec θ + tan θ| + C = ln|x + x2 − 1| + C
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #12
READY,
GET SET,…
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
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INTEGRAL #12
∫
sin2 x cos2 x dx
2 : 30
2 0 1 3
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #12
∫
sin2 x cos2 x dx
∫(
)
1 + cos 2x
=
dx
2
)
∫
∫
∫(
(
)
1
1
1 − cos 4x
1
1 − cos2 2x dx =
sin2 2x dx =
=
4
4
4
2
1 − cos 2x
2
)(
(
)
1
1
sin 4x
=
(1 − cos 4x) dx =
x−
+C
8
8
4
∫
2 0 1 3
U
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