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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