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Millersville University Name Answer Key
Department of Mathematics
MATH 211, Calculus II, Homework 1
January 24, 2008
Please answer the following questions. Partial credit will be given as appropriate, do not
leave any problem blank. Each problem is worth ten points. Your completed assignment
will be due at class time on Monday, January 28, 2008.
- Evaluate the following definite and indefinite integrals.
(a)
5 โ x โ 25 โ x^2
dx
5 โ x โ 25 โ x^2
dx =
25 โ x^2
dx โ
x โ 25 โ x^2
dx
x 5
) 2 dx^ โ
x โ 25 โ x^2
dx
x 5
) 2 dx^ โ
x โ 25 โ x^2
dx
In the first integral make the substitution
u =
x
5
5 du = dx
and in the second integral make the substitution
w = 25 โ x
2 โ
dw = x dx.
5 โ x โ 25 โ x^2
dx =
1 โ u^2
du โ
w
โ 1 / 2 dw
1 โ u^2
du +
w
โ 1 / 2 dw
= 5 arcsin u + w
1 / 2
= 5 arcsin
x
5
25 โ x^2 + C
(b)
(x + 4)^2 + 1
dx
Make the substitution
u = x + 4 du = dx.
(x + 4)^2 + 1
dx =
u^2 + 1
du
= arctan u + C
= arctan(x + 4) + C
(c)
cos x โ 1 + sin
2 x
dx
Make the substitution
u = sin x du = cos x dx.
cos x โ 1 + sin
2 x
dx =
1 + u^2
du
= sinh
โ 1 u + C
= sinh
โ 1 (sin x) + C
(d)
x tan x sec x dx
Integrate by parts letting
u = x
du = dx
v = sec x
dv = sec x tan x dx
x tan x sec x dx = x sec x โ
sec x dx
= x sec x โ
sec x
sec x + tan x
sec x + tan x
dx
= x sec x โ
sec^2 x + sec x tan x
sec x + tan x
dx
Make the substitution
u = sec x + tan x du = (sec^2 x + sec x tan x) dx.
x tan x sec x dx = x sec x โ
u
du
= x sec x โ ln |u| + C
= x sec x โ ln | sec x + tan x| + C
- Differentiate the following functions.
Thus
lim xโ 0
arcsin x ฯ 2 โ^ arccos^ x^
(b) lim xโโ
ฯ
2
โ arctan x
cot
x
Since limxโโ arctan x = ฯ/2 and limxโโ cot(1/x) = limzโ 0 +^ cot z = โ the
original limit is indeterminate of the form 0 ยท โ. We can see that
lim xโโ
ฯ
2
โ arctan x
cot
x
= lim xโโ
ฯ 2 โ^ arctan^ x tan
1 x
where the limit on the right-hand side of the equation is indeterminate of the form
0 /0. We may apply lโHหopitalโs rule.
lim xโโ
d dx
ฯ 2 โ^ arctan^ x
d dx
tan
1 x
) = lim xโโ
1 1+x^2
โ (^) x^12 sec^2
1 x
= lim xโโ
1 1+x^2 1 x^2 sec
2
1 x
= lim xโโ
x^2
1 + x^2
cos
2
x
[
lim xโโ
x^2
1 + x^2
] [
lim xโโ
cos
2
x
)]
= lim xโโ
x^2
1 + x^2
= lim xโโ
1 x^2 + 1 = 1
Thus
lim xโโ
ฯ
2
โ arctan x
cot
x
(c) lim xโ 1 +
cosh
โ 1 x
x โ 1
Since limxโ 1 + cosh
โ 1 x = 0 and limxโ 1 + (x โ 1) = 0 the original limit is indeter-
minate of the form 0/0. We can apply lโHหopitalโs rule.
lim xโ 1 +
d dx
cosh
โ 1 x
d dx (x^ โ^ 1)^
= lim xโ 1 +
โ^1 x^2 โ 1
1
= lim xโ 1 +
x^2 โ 1 = โ
Therefore we have
lim xโ 1 +
cosh
โ 1 x
x โ 1
