Calc #2 Practice Test #4 - SOLUTIONS
Summer 1997
Pat Rossi
Name
1. ~Write the given expressions in algebraic form.
(a) cos(arcsin (2x)) =
Let y = arcsin (2x)
This is the same as saying “y is the angle whose sine is 2x.”
i.e., sin (y) = 2x.
Let’s draw a triangle to depict this relationship.
1
2x
y
ÓÓÓÓÓ
p 1 ! 4x2
By q
Pythagorean’s Theorem, (opp)2 + (adj)2 = (hyp)2
√
⇒ (hyp)2 − (opp)2 = adj ⇒ adj = 1 − 4x2 .
Anyway, we want cos(arcsin (2x)) = cos (y)
√
1 − 4x2
¡
¢
(b) tan (tan−1 (x)) = x for all values of x in the interval − π2 , π2 . This is by definition
of tan−1 (x).
From the picture, cos (y) =
adj
hyp
=
√
1−4x2
1
=
(c) sec (tan−1 (x)) =
Let y = tan−1 (x)
This is the same as saying “y is the angle whose tangent is x.”
i.e., tan (y) = x.
Let’s draw a triangle to depict this relationship.
x
y
1
By Pythagorean’s
Theorem, (opp)2 + (adj)2 = (hyp)2 ⇒ 12 + x2 = (hyp)2 ⇒
√
2
(hyp) = 1 + x2
Anyway, we want sec (tan−1 (x)) = sec (y) .
√
√
1+x2
From the picture, sec (y) = hyp
=
=
1 + x2 .
adj
1
¡
¡
¢¢
(d) cos arcsin x−h
=
r¡
¢
x−h
Let y = arcsin r
.”
This is the same as saying “y is the angle whose sine is x−h
r
x−h
i.e., sin (y) = r .
Let’s draw a triangle to depict this relationship.
r
x!h
ÓÓÓÓÓÓÓ
p r2 ! (x ! h)2
y
By q
Pythagorean’s Theorem, (opp)2 +q
(adj)2 = (hyp)2
(hyp)2 − (opp)2 = adj ⇒ adj = r2 − (x − h)2 .
√
¡
¡ x−h ¢¢
r2 −(x−h)2
adj
= cos (y) = hyp =
Anyway, we want cos arcsin r
r
⇒
2.
d
1
[arccos (3x − π)] = − q
· |{z}
3 = −√ 3
1−(3x−π)2
{z
}
|dx
1 − (3x − π)2 du
d
[arccos(u)]
|
{z
} dx
dx
1
1−u2
−√
3.
d
dx
£ −1 √ ¤
d h −1 ³ 1 ´i
1
1 1
= r
· x− 2 =
sin
x2
sin ( x) =
³
´
1 2
{z
}
|2 {z }
|dx
1 − x2
d
du
[arcsin(u)]
dx
dx
|
{z
}
1
1−u2
√
√1
2 x−x2
4.
¡ ¢¤
d £
1
arctan 3x2 =
· 6x =
1 + (3x2 )2 |{z}
{z
}
|dx
| {z } du
dx
d
dx
5.
R1
0
[arctan(u)]
√ 1
dx
4−x2
6x
1+9x4
1
1+u2
=
This fits the form:
R
√ 1
du
a2 −u2
= arcsin
¡u¢
a
2
+C
1
1√
2x 2 1−x
=
1
√ √
2 x 1−x
=
a2 = 4
⇒a=2
u2 = x2
Here, ⇒ u = x
⇒ du = dx
when x = 0; u = x = 0
when x = 1; u = x = 1
£
¡ u ¢¤u=1
£
¡ u ¢¤u=1
R x=1 1
R u=1 1
√
Therefore, x=0 √4−x
dx
=
dx
=
arcsin
=
arcsin
=
2
a u=0
2 u=0
u=0
a2 −u2
¡1¢
π
π
arcsin 2 = arcsin (0) = 6 − 0 = 6
R e2x
R
1
6. 4+e
e2x dx
4x dx =
22 +(e2x )2
¡u¢
R 1
1
This fits the form: a2 +u
+C
2 du = a arctan a
a2 = 4
⇒a=2
u2 = e4x
⇒ u = e2x
⇒ du = 2e2x dx
⇒ 12 du = e2x dx
Z
R e2x
R
1
Therefore,
dx =
e|2x{zdx} =
2
4+e4x
22 + (e2x ) 1
| R {z
} 2 du
1
a2 +u2
· 12 du =
1
1
a
7.
arctan
R2
√2
3
¡u¢
a
+C =
√1
dx
x x2 −1
1
2
=
This fits the form:
R
1
2
· arctan
+u2
´
³a22x
√ 1
du
u u2 −a2
e
2
=
1
a
1
4
+ C = arctan
arcsec
³ ´
|u|
a
³
e2x
2
´
1
2
R
1
du
a2 +u2
+C =
+C
a2 = 1
⇒a=1
u2 = x2
⇒u=x
⇒ du = dx
when x = √23 ; u = x = √23
when x = 2; u = x = 2
h
³ ´iu=2
R x=2
R u=2
Therefore: x= √2 x√x12 −1 dx = u= √2 u√u12 −a2 du = a1 arcsec |u|
=
a
3
3
u= √2
3
³ ´
arcsec (2) − arcsec √23 = π3 − π6 = π6
Remark 1 How did I figure the values of arcsec (2) and arcsec
¡ ¢
¡ ¢
Observe: cos π3 = 12 ⇒ sec π3 = 2 ⇒ arcsec (2) = π3
3
³
√2
3
´
?
=
1
2
·
Also: cos
¡π¢
6
=
√
3
2
⇒ sec
¡π¢
6
=
³ √1 ´
3
2
=
4
√2
3
⇒ arcsec
³
√2
3
´
=
π
6