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Calculus II Test
#3 April
15, 2005
Name____________________ R. Hammack Score
______
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(1) ∫ dx
=
∫ cos(
)2x
dx
=
sin(
)
+ C
(2) ∫ x
(x) dx
= x tan(x)-∫ tan(x) dx
= x tan(x)-ln| sec(x) | + C
(3) ∫
sec(x) dx
= ln| sec(x) + tan(x) | + C
(4)
(x)dx
=
x (x)
-
x
dx
=
x (x)
+
(-2x
) dx
=
x (x)
+
+
C =
x (x)
+
+
C
(Using interation by parts with the following
substitutions:)
(5) ∫
dx
= -2∫
dx
= -2∫
du
=
+ C =
+ C =
+
C
(6) ∫
(x)
(x)
dx
= ∫
(x)
(x)cos(x)
dx
=
∫ (x)
(1-
(x))cos(x)
dx
= ∫
(1-
)du
=∫
-
du
=
-
+C=
-
+
C
(7) ∫
dx
= ∫
dx
=
∫ du
=
(
)
+ C =
(
)
+ C
(8)
dx
=
dx
=
=
(1)
=
(9) ∫
dx
= ∫
+
dx
=
∫
dx
+ ∫
dx
+ ∫
dx
=
=ln(
+2)
+
(x)
-
+
C
A =1
B =1
C = 2
D = 0
(10) ∫
dx
=∫
sec(θ)tan(θ) dx
=
∫ dx
= ∫sec(θ)
dx
= ln| sec(θ) + tan(θ) | + C
=ln| x+tan((x))
|+C=ln| x +
|
+ C
(11)
dx
=
dx =
=
=
(-
+
)
= 0 +
=
(12)
dx
=
(Note: The integral is improper)
=3
dx =
=
=
(6
-6
)
= 6
(13)
=
=
(14) x
=
=
= 0