Name: Student ID #: JUSTIFY ALL YOUR ANSWERS

MATH 206
Calculus II
Exam 2
Name:
Friday, October 17, 2003
Student ID #:
JUSTIFY ALL YOUR ANSWERS
DO NOT ANSWER YOUR QUESTIONS ON THIS PIECE OF PAPER
USE PENCIL
This exam has 5 questions, for a total of 80 points.
1. Compute the following integrals.
(a) (10 points)
Z
π/12
tan 3x dx;
0
(b) (10 points)
(c) (10 points)
Z
Z
dx
;
x (log8 x)2
π/2
2 sinh (sin θ) cos θdx;
0
(d) (10 points)
Z
1
2
x2
8dx
.
− 2x + 2
2. (10 points) Use logarithmic differentiation to compute the derivative of y = (ln x)ln x
with respect to x.
3. (10 points) Use implicit differentiation to compute dy/dx for tan y = ex + ln x.
4. (10 points) Deduce the formula for the derivative of y = tan−1 x with respect to x.
√
dy
5. (10 points) Solve the differential equation 2xy dx
= 1.
1
x
1
csch−1 = sinh−1
x
1
coth−1 = tanh−1
x
sinh 2x = 2 sinh x cosh x
sech−1 = cosh−1
cosh 2x = cosh2 x + sinh2 x
cosh 2x + 1
2
cosh 2x − 1
sinh2 x =
2
cosh2 x =
2
d(sinh−1 u)
1
du
=√
dx
1 + u2 dx
−1
1
du
d(cosh u)
=√
,u>1
dx
u2 − 1 dx
d(tanh−1 u)
1 du
=
, |u| < 1
dx
1 − u2 dx
d(coth−1 u)
1 du
=
, |u| > 1
dx
1 − u2 dx
−du/dx
d(sech−1 u)
= √
,0<u<1
dx
u 1 − u2
−du/dx
d(csch−1 u)
√
=
, u 6= 0
dx
|u| 1 + u2
2
cosh x − sinh x = 1
tanh2 x = 1 − sech2 x
coth2 x = 1 + csch2 x
d
du
(sinh u) = cosh u
dx
dx
du
d
(cosh u) = sinh u
dx
dx
d
du
(tanh u) = sech2 u
dx
dx
du
d
(coth u) = − csch2 u
dx
dx
d
(sech u) = − sech u tanh u
dx
d
(csch u) = − csch u coth u
dx
Z
du
dx
du
dx
Z
sinh u du = cosh u + C
Z
cosh u du = sinh u + C
Z
Z
sech2 u du = tanh u + C
csch2 u du = − coth u + C
Z
sech u tanh u du = − sech u + C
Z
csch u coth u du = − csch u + C
du
−1 u
+ C, a > 0
= sinh
a
a2 + u 2
Z
u
du
√
= cosh−1
+ C, u > a > 0
a
u 2 − a2 
−1 u
1
Z
tanh
+ C, u2 < a2

a
a
du
=
a2 − u2  1
−1 u
coth
+ C, u2 > a2
a
a
Z
1
du
u
√
= − sech−1
+ C, 0 < u < a
2 − u2
a
a
u
a
Z
u
1
du
√
= − csch−1 + C, u 6= 0
2
2
a
a
u a +u
√