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Quiz 3
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||||||||||||||||||||||||||||||||||||||||||||||||||||||1 (4 points) First make a substitution and then use integration by parts to evaluate the integral
p
1
dx and dx = 2udu.
Let u = x. Then du = 2p1 x dx = 2u
Z
p
cos xdx =
Z
Z
R
p
cos xdx.
Z
u d(sin u)2u sin u ¡ 2 sin u du
p
p
p
= 2u sin u + 2 cos u + C = 2 x sin x + 2 cos x + C:
R
2 (4 points) (extra 2 points if using two correct methods) Evaluate the integral sin(2x) cos(2x)dx
Method 1) sin(2x) cos(2x) =
1
2
(cos u)2u du = 2
sin(4x).
Z
sin(2x) cos(2x)dx =
1
2
Z
1
sin(4x)dx = ¡ cos(4x) + C:
8
Method 2) Let u = sin(2x). Then du = 2 cos(2x)dx and cos(2x)dx = 12 du.
Z
Z
1
1
1
sin(2x) cos(2x) dx =
udu = u2 + C = sin2 (2x) + C:
2
4
4
3 (2 points) Evaluate the integral
Z
1
p
dx. Express your answer without the inverse of any trigonometric functions.
x2 4 ¡ x2
Let x = 2 sin µ. Then dx = 2 cos µ dµ.
Z
Z
Z
2 cos µ
1
1
1
p
dx
=
dµ
=
dµ
4
4 sin2 µ ¢ 2 cos µ
sin2 µ
x2 4 ¡ x2
p
p
Z
1
1 1 ¡ (x=2)2
1
4 ¡ x2
2
=
+C =¡
+ C:
csc µdµ = ¡ cot µ + C = ¡
4
4
4
(x=2)
4x
How to ¯nd cot µ ? Take a point P (a; b) on the terminal side of the angle µ. By de¯nition sin µ = rb , where r =
p
p
We can take r = 1 and b = x2 , since sin µ = x2 . Observe that a = r2 ¡ b2 = 1 ¡ (x=2)2 . Thus
a
cot µ = =
b
p
1
1 ¡ (x=2)2
:
(x=2)
p
a2 + b2 .