Calculus II
Math 116
Homework 2
Due Friday Feb. 7
(1) Compute the derivatives of each of the following functions:
(a) f (x) = cos2 (x).
Solution. f (x) = cos2 (x) = cos(x) cos(x).
f 0 (x) = − sin(x) cos(x) − cos(x) sin(x) = −2 sin(x) cos(x).
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x2
(b) g(x) = sin(e )
2
2
Solution. g 0 (x) = 2xex cos(ex ).
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(c) h(x) = sin(tan(x))
Solution. h0 (t) = sec2 (x) cos(tan(x)).
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(d) k(x) = ln(csc(x))
Solution. k 0 (x) =
− csc(x) cot(x)
csc(x)
= − cot(x).
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(2) Sketch the graph of the function f (x) = x−cos(x) over the interval [0, 2π] by obtaining
the following information:
See graph on last page.
(a) The intervals where f is increasing and where it is decreasing
Solution. f 0 (x) = 1 + sin(x). Then f 0 (x) = 0 when sin(x) = −1. So f 0 (x) = 0
when x = 3π
. Then we break up the domain of f , [0, 2π], into [0, 3π/2) and
2
0
(3π/2, 2π]. f is positive on both [0, 3π/2) and (3π/2, 2π].Thus f is increasing
over [0, 2π].
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(b) The relative extrema of f
Solution. Either f has a no relative extrema since f is increasing over its entire
domain or the relative extrema occur at the endpoints.
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(c) The concavity of f
Solution. f 00 (x) = cos(x). f 00 (x) = 0 when x = π2 and x = 3π
. Then we break
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up the domain of f into [0, π/2), (π/2, 3π/2) and (3π/2, 2π]. f 00 is positive on
[0, π/2) and (3π/2, 2π], and f 00 is negative on (π/2, 3π/2). Thus f is concave up
on[0, π/2) and (3π/2, 2π], and concave down on (π/2, 3π/2).
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(d) The inflection points of f
Solution. f has inflection points when x =
π
2
and x =
(3) FindR the indefinite integral:
(a) (cos(x) + 4)dx
R
Solution. (cos(x) + 4)dx = sin(x) + 4x + C.
R
(b) (x8 + 3 sin(x))dx
R
Solution. (x8 + 3 sin(x))dx = 91 x9 − 3 cos(x) + C.
R
√
(c) ( x3 − x + sec2 (x))dx
1
3π
.
2
4
4
4
2
R
√
3
Solution. ( x3 − x + sec2 (x))dx = 3 ln |x| − 32 x 2 + tan(x) + C.
R
(d) (5ex + 2)dx
R
Solution. (5ex + 2)dx = 5ex + 2x + C.
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(4) Use Rthe substitution method to find the indefinite integral:
(a) 4x(2x2 − 5)3 dx
Solution. Let u = 2x2 − 5. Then du = 4xdx.
Z
Z
2
3
4x(2x − 5) dx = u3 du = 41 u4 + C = 14 (2x2 − 5)4 + C.
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(b)
R
x2 +1
x3 +3x+2
dx
Solution. Let u = x3 + 3x + 2. Then du = (3x2 + 3)dx = 3(x2 + 1)dx.
Z
Z
x2 + 1
1
1
dx = 3
dx = 13 ln |u| + C = 13 ln |x3 + 3x + 2| + C.
3
x + 3x + 2
u
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(c)
R
sin(tan(x))
dx
cos2 (x)
Solution. Let u = tan(x). Then du = sec2 (x)dx = cos12 (x) dx.
Z
Z
sin(tan(x))
dx = sin(u)du = − cos(u) + C = − cos(tan(x)) + C.
cos2 (x)
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(d)
R
esin(x) cos(x)dx
Solution. Let u = sin(x). Then du = cos(x)dx.
Z
Z
sin(x)
e
cos(x)dx = eu du = eu + C = esin(x) + C.
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(5) The velocity, in feet per second, of a rocket t seconds into vertical flight is given by
v(t) = −3t2 + 192t
Find an expression h(t) that gives the rocket’s height, in feet, t seconds after liftoff.
You can assume that the height of the rocket at time t = 0 is 0 feet.
Solution.
Z
h(t) =
Z
v(t)dt =
(−3t2 + 192t)dt = −t3 + 96t2 + C.
Since h(0) = 0, that is, the rocket took off from the ground, we have h(0) = C = 0.
Thus h(t) = −t3 + 96t2 .
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3
(6) Evaluate the limit:
cos(x) − 1
.
x→0
x
lim
Solution.
0
cos(x) − 1 0
− sin(x)
= 0.
= lim
x→0
x→0
x
1
lim
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* (extra credit) Using the substitution method compute:
Z
csc(x)dx.
Hint: First multiply the numerator and denominator of the integrand by
(csc(x) − cot(x)).
Solution.
Z
Z Z csc(x) − cot(x)
csc(x)(csc(x) − cot(x))
csc(x)dx =
csc(x) ·
dx =
dx
csc(x) − cot(x)
csc(x) − cot(x)
Z 2
csc (x) − csc(x) cot(x)
=
dx.
csc(x) − cot(x)
Now let u = csc(x) − cot(x). Then du = (− csc(x) cot(u) + cot(x))dx. Thus
Z
Z 2
csc (x) − csc(x) cot(x)
1
du = ln |u| + C.
dx =
csc(x) − cot(x)
u
So
Z
csc(x)dx = ln | csc(x) − cot(x)| + C.
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