Math 2300: Calculus II
Tables, Computer Algebra Systems, and alternate solutions
Even when we stay within the realm of elementary functions, the general families of integrals that
come up can be quite varied. So when integrating, it will often be practical to use a table of integrals
or a computer algebra system (Wolfram Alpha, Maple, Mathematica, etc.), rather than work out an
integral by hand. However, antiderivatives are only unique up to a constant, so different methods
may produce answers that look quite different.
Z
dx
√
1. In this question we will investigate the integral
.
9 + x2
(a) Compute the antiderivative using a trig substitution.
We will make a substitution using tan x since we see the form a2 + x2 :
(
Z
Z
x = 3 tan θ
dx
3 sec2 θdθ
√
√
.
=⇒
dx = 3 sec2 θdθ
9 + x2
9 + 9 tan2 θ
Z
Z
Z
3 sec2 θdθ
sec2 θdθ
√
√
Now,
=
= sec θdθ = ln|sec θ + tan θ| + C.
3 1 + tan2 θ
sec2 θ
x
Don’t forget to substitute back θ = arctan
, for example by using the triangle:
3
√
x
9 + x2
Solution:
θ
Z
This give us the final answer
3 √9 + x2 x dx
√
= ln + + C.
2
3
3
9+x
(b) What is the antiderivative according to the table of integrals in your book? Can you
make the two answers agree?
Solution:
From the integral table in the text, this integral matches number 25,
Z
√
R dx
du
2 + C.
√
with a = 3 and u = x. Hence √9+x
=
ln
x
+
9
+
x
2
a2 + x2
Simplifying the answer
from part b:
!
√
p
p
2
x
9+x
ln
+
+ C = ln x + 9 + x2 − ln 3 + C = ln x + 9 + x2 + C 0 .
3
3
The term inside the natural log must always be positive, so the absolute value signs are
not necessary.
(c) What does Wolfram Alpha say this integral is? Is Wolfram wrong?
Z
x
dx
= sinh−1
+C, which clearly
3
9 + x2
looks different than the answers we have come up with above. What is sinh−1 ?
Solution:
Wolfram Alpha gives an answer
1
√
Math 2300: Calculus II
Tables, Computer Algebra Systems, and alternate solutions
2. The function sinh x, pronounced “cinch x”, is the hyperbolic sine function. So sinh−1 x (also
known as arcsinhx) is the inverse of sinh x.
(a) Use the internet to determine the definition of sinh x in terms of exponential functions.
Solution:
ex − e−x
sinh x :=
2
(b) Use the internet to determine the logarithmic representation of sinh−1 x.
Solution:
p
sinh−1 x := ln x + x2 + 1 .
This identity can be proven by using the definition of an inverse function.
ey − e−y
for y).
x=
2
(Solve
(c) Verify that your answers to 1(a) and 1(b) agree with Wolfram Alpha’s answer.
Solution: Given the new identity, we find that all three answers from above do in fact
agree up to a constant, as
!
r
p
x
x2
−1 x
+ C = ln
+
+ 1 + C = ln x + 9 + x2 − ln(3) + C.
sinh
3
3
9
Note there is something to be said about
domain, as the answer we get by hand includes
√
absolute value bars. However, x + 9 + x2 is always positive, and the three answers do
indeed agree.
Z
3. Compute
1
dx:
x2 − 1
(a) Using partial fractions
Solution: We first must find the partial fraction decomposition.
1
A
B
=
+
x2 − 1
x−1 x+1
1 = A(x + 1) + B(x − 1)
1 = A − B + (A + B)x
(
A−B =1
A+B =0
=⇒ A = −B =
1
2
Now the problem has been simplified to the integral
Z
Z
Z
1/2
1/2
1
1
1
−
dx =
dx −
dx
x−1 x+1
2
x−1
x+1
1
1
= ln|x − 1| − ln|x + 1| + C
2
2
2
Math 2300: Calculus II
Tables, Computer Algebra Systems, and alternate solutions
(b) Using the tables in your book. Then show that the two answers actually agree.
Z
du
Solution: Number 20.
matches with a = 1 and u = x, so we have
2
u − a2
Z
dx
1 x − 1 1
1
=
ln
+ C = ln|x − 1| − ln|x + 1| + C.
x2 − 1
2 x + 1
2
2
(c) We can also approach this problem with a trig substitution. We see the form x2 − a2 ,
and looking at the pythagorean identity tan2 θ = sec2 θ − 1 suggests that we could use a
secant substitution.
(
Z
Z
x = sec θ
dx
sec θ tan θdθ
=⇒
2
x −1
sec2 θ − 1
dx = sec θ tan θdθ
Z
Z
sec θdθ
sec θ tan θdθ
=
=
2
tan θ
tan θ
Z
= csc θdθ = − ln |csc θ + cot θ| + C
Now complete the integral by using the triangle below to substitute back θ =
√
x
θ
arcsec(x)
x2 − 1
1
x+1 x
1
+C =
+ C = − ln √
Solution: − ln |csc θ + cot θ| + C = − ln √
+√
x2 − 1
x2 − 1 x2 − 1 √
x2 − 1 ln +C
x+1 (d) Use algebraic manipulation to show that this third answer agrees with the answers from
parts (a) and (b).
Solution:
√
r
√
√
x2 − 1 x − 1
x + 1 x − 1
1 x − 1 ln + C = ln + C = ln x + 1 + C = 2 ln x + 1 + C
x+1 x+1
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Math 2300: Calculus II
Tables, Computer Algebra Systems, and alternate solutions
4. Find the number of the form in the table of integrals in your book that you can use to find
each of the following integrals, and give the correct antiderivative.
Z
x2 + 25
√
(a)
dx
x2 + 25
Z p
p
up 2
a2 a2 + u2 du =
a + u2 +
ln u + a2 + u2 + C
2
Z p
Z 22
x + 25
√
dx =
We get to this form by simplifying the ratio,
x2 + 25dx. Thus we
x2 + 25
use the table with a = 5 and u = x to get
Z
p
x2 + 25
xp
25 √
dx =
25 + x2 +
ln x + 25 + x2 + C
2
2
x2 + 25
Solution:
21.
(b) Use the table of integrals to inspire a choice for a u/du
substitution. Make the substiZ
csc2 (3x)dx
p
tution, then complete the integral using the table.
12 + tan2 (3x)
√
Solution: The integral tables have lots of entries with a2 + x2 in them. This inspires
the
Z substitution u = tan(3x). We use some trig identities to rewrite the integral as
1
du
√
.
2
3
u 12 + u2
Next we find the right line
√ of the table:
Z
du
a2 + u2
√
28.
=−
+C
a2 u
u2 a2 + u2
Using this fact, we get: p
Z
csc2 (3x)dx
12 + tan2 (3x)
p
=−
+C
36 tan(3x)
12 + tan2 (3x)
We can see that using tables of integrals, as well as interpreting the results of a computer
algebra system integration requires the ability to recognize and rewrite functions in various
forms. If you find yourself using these methods to compute an integral, just remember that
with a little work and perhaps some research, you can make sense of an unfamiliar answer!
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