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 attempted 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 2
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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 2
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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
∫
1 + sin x
dx
cos x
2 : 30
2 0 1 2
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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
∫
1 + sin x
dx
cos x
∫(
=
1
sin x
+
cos x cos x
)
dx
∫
= (sec x + tan x) dx
= ln|sec x + tan x| + ln|sec x| + C
2 0 1 2
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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 2
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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
dx
4
x +1
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #2
∫
x
dx
4
x +1
1
=
2
∫
[
1
2
d
u
u
=
x
,
u2 + 1
du = 2x dx
]
1
= arctan u + C
2
( 2)
arctan x
+C
=
2
2 0 1 2
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 2
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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 #3
∫
sin4 x cos3 x dx
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #3
∫
sin4 x cos3 x dx
∫
=
∫
sin4 x cos2 x cos x dx =
∫1
=
(
4
6
u −u
)
du
sin4 x(1 − sin2 x) cos x dx
[ u = sin x,
du = cos x dx ]
0
=
sin5 x
2 0 1 2
5
−
U
sin7 x
7
of
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+C
I N T E G R A T I O N
B E E
INTEGRAL #4
READY,
GET SET,…
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
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INTEGRAL #4
∫
ln ln x
dx
x ln x
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #4
∫
ln ln x
dx
x ln x
[
∫
=
u du
u = ln ln x, du =
1
x ln x
]
dx
u2
=
+C
2
(ln ln x)2
+C
=
2
2 0 1 2
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 2
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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 #5
∫
x+1
dx
2
x +1
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #5
∫
x+1
dx
2
x +1
∫(
=
=
x
1
+
x2 + 1 x2 + 1
ln(x2 + 1)
2 0 1 2
2
U
of
)
dx
+ arctan x + 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 2
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I N T E G R A T I O N
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INTEGRAL #6
∫
sec x tan5 x dx
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #6
∫
sec x tan5 x dx
∫
∫
tan4 x · sec x tan x dx = (sec2 x − 1)2 · sec x tan x dx
=
∫
= (u2 − 1)2 du [ u = sec x,
∫
4
2
= (u − 2u + 1) du =
2 0 1 2
U
of
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sec5 x
5
du = sec x tan x dx ]
2 sec3 x
−
+ sec x + C
3
I N T E G R A T I O N
B E E
INTEGRAL #7
READY,
GET SET,…
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
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INTEGRAL #7
∫
x2
√
1
1 − x2
dx
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #7
∫
x2
√
[
1
1 − x2
dx
x = sin θ,
∫
=
∫
=
]
√
1 − x2 = cos θ
dx = cos θ dθ,
1
· cos θ dθ =
2
sin θ cos θ
∫
1
sin2 θ
dθ
√
1 − x2
+C
csc θ dθ = − cot θ + C = −
x
2 0 1 2
2
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 2
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I N T E G R A T I O N
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INTEGRAL #8
∫
x
dx
2
x +x−6
2 : 30
2 0 1 2
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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
∫
x
dx
2
x +x−6
∫(
=
3
1
2
1
·
+ ·
5 x+3 5 x−2
)
dx
[ partial fractions ]
3
2
= ln|x + 3| + ln|x − 2| + C
5
5
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #9
READY,
GET SET,…
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #9
∫
ln tan x
dx
sin x cos x
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #9
∫
ln tan x
dx
sin x cos x
[
2
u = ln tan x,
du =
sec x
1
dx =
dx
tan x
sin x cos x
]
∫
=
u du
u2
(ln tan x)2
=
+C=
+C
2
2
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #10
READY,
GET SET,…
2 : 30
2 0 1 2
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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 #10
∫
1 − etan x
dx
2
1 − sin x
2 : 30
2 0 1 2
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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
∫
1 − etan x
dx
2
1 − sin x
∫
=
∫
1 − etan x
tan x
2
(1
−
e
)
sec
x dx
d
x
=
cos2 x
∫
u
= (1 − e ) du
[
u = tan x,
2
du = sec x dx
]
= tan x − etan x + C
2 0 1 2
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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
READY,
GET SET,…
2 : 30
2 0 1 2
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I N T E G R A T I O N
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INTEGRAL #11
∫
1
dx
√ √
x 1−x
2 : 30
2 0 1 2
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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
√ √
x 1−x
∫
=2 √
1
du
2
1−u
[
u=
√
x,
1
]
du = √ dx
2 x
= 2 arcsin u + C
√
= 2 arcsin x + C
2 0 1 2
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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 #12
READY,
GET SET,…
2 : 30
2 0 1 2
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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 #12
∫
ln x
2
x + x ln x
dx
2 : 30
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #12
∫
ln x
2
x + x ln x
∫
=
dx
ln x
dx
2
x(1 + ln x)
[
∫
1 1
=
du
u = 1 + ln2 x,
2 u
(
2 ln x
du =
dx
x
]
)
ln 1 + ln x
ln u
=
+C=
+C
2
2
2 0 1 2
U
of
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2
I N T E G R A T I O N
B E E
INTEGRAL #13
READY,
GET SET,…
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #13
∫
cos 2x (sin x + cos x)2 dx
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #13
∫
cos 2x (sin x + cos x)2 dx
∫
=
(
)
cos 2x sin x + 2 sin x cos x + cos x dx
2
2
∫
=
∫
cos 2x (1 + 2 sin x cos x) dx =
(1 + sin x)2
=
4
2 0 1 2
U
of
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or
sin 2x
2
+
cos 2x (1 + sin 2x) dx
sin2 2x
4
or
sin 2x
2
I N T E G R A T I O N
−
cos2 2x
4
B E E
INTEGRAL #14
READY,
GET SET,…
2 : 30
2 0 1 2
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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 #14
∫
1
dx
(x − 3)(x + 1) + 5
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #14
∫
1
dx
(x − 3)(x + 1) + 5
∫
=
∫
=
1
dx
x2 − 2x + 2
1
dx
(x − 1)2 + 1
= arctan(x − 1) + C
2 0 1 2
U
of
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I N T E G R A T I O N
B E E
INTEGRAL #15
READY,
GET SET,…
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #15
∫
√
ln x2 + 1 dx
2 : 30
2 0 1 2
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of
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I N T E G R A T I O N
B E E
INTEGRAL #15
∫
√
ln x2 + 1 dx
2
∫
x2
dx [ integrate by parts ]
2
x +1
)
∫(
2
x ln(x + 1)
1
=
−
1− 2
dx
2
x +1
x ln(x + 1)
=
−
2
x ln(x2 + 1)
=
− x + arctan x + C
2
2 0 1 2
U
of
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I N T E G R A T I O N
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