Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
6.4
Integral Tables
If the standard integration techniques presented previously fail to yield an
antiderivative, the last measures of despair are integral tables. These tables
basically consist of collections of functions together with their antiderivatives. In order to use them, you may have to re-write the integrand function
first in a standard form listed in the table.
Example 6.4.1
R
Evaluate, using the tables of integrals:
cos x
dx.
sin2 x−9
Solution.
Letting u = sin x so that du = cos xdx, the given integral converts to
Z
Z
cos x
du
.
dx =
2
2
u
−9
sin x − 9
Using Formula (20), we find
Z
Z
du
cos x
dx
=
2
2
u −9
sin x − 9
1 u − 3 = ln +C
6
u + 3
1 sin x − 3 = ln +C
6
sin x + 3 Example
6.4.2
R
Find sin4 xdx.
Solution.
Using the reduction formula (73) twice, first with n = 4 and then with n = 2,
we find
Z
Z
1
3
4
3
sin xdx = − sin x cos x +
sin2 xdx
4
4
1
3
3
= − sin3 x cos x − sin x cos x + x + C
4
8
8
1
Example 6.4.3
R √
Using the tables of integrals, evaluate: x 6 + 4x − 4x2 dx.
Solution.
We have
Z p
Z
p
1
2
x 6 + 4x − 4x dx = −
(4 − 8x − 4) 6 + 4x − 4x2 dx
8
Z
Z
p
1 p
1
2
(4 − 8x) 6 + 4x − 4x dx +
6 + 4x − 4x2 dx
=−
8
2
Z
Z
1 √
1 p
=−
udu +
7 − w2 dw
8
4
3
1 u2
1 wp
7
−1 w
2
=− 3 +
7 − w + sin √ + C
8 2
4 2
2
7
p
3
1
2x
−
1
= − (6 + 4x − 4x2 ) 2 +
6 + 4x − 4x2
12 8
7
2x − 1
√
+C
+ sin−1
8
7
where u = 6 + 4x − 4x2 and w = 2x − 1. For the integral in w, we used
formula (30)
Example 6.4.4
R√
Using the tables of integrals, evaluate:
e2x − 1dx.
Solution.
Using the substitution u = ex we find du = ex dx = udx so that dx =
Hence, using this substitution and formula (41), we find
Z p
Z √ 2
u −1
e2x − 1dx =
du
u
p
1
−1
2
= u − 1 − cos
+C
ex
p
= e2x − 1 − cos−1 (e−x ) + C
Example 6.4.5
R
3
Find tan x(1/x)
dx.
2
2
du
u .
Solution.
Using the substitution u = x1 and formula (69), we find
Z
Z
tan3 (1/x)
dx = − tan3 udu
x2
1
= − tan2 u − ln | cos u| + C
2
1
= − tan2 (1/x) − ln | cos (1/x)| + C
2
Example
6.4.6 (Completing the square)
R
1
Find x2 +4x+5
dx.
Solution.
Completing the square we find x2 + 4x + 5 = (x + 2)2 + 1. Now, letting
u = x + 2 we can write
Z
Z
1
1
dx
=
du.
2
2
x + 4x + 5
u +1
Finally, using formula (17) with a = 1 we find
Z
Z
1
1
dx =
du = arctan u + C = arctan (x + 2) + C
2
2
x + 4x + 5
u +1
Example R6.4.7
4
Evaluate: √xx10 −1 dx.
Solution.
Using the substitution u = x5 and formula (43), we find
Z
Z
x4
du
√
dx = √
du
10
x −1
u2 − 1
p
= ln |u + u2 − 1| + C
p
= ln |x5 + x10 − 1| + C
Example
6.4.8 (Using substitution)
R
2
Find xe2x cos (2x2 )dx.
Solution.
Let u = 2x2 . Then du = 4xdx so that
Z
Z
1
2x2
2
xe cos (2x )dx =
eu cos udu.
4
3
Now, applying formula (99) to the last integral, with a = 1 and b = 1, we
find
Z
1
eu cos udu = eu (cos u + sin u) + C.
2
Hence,
Z
1 2
2
xe2x cos (2x2 )dx = e2x (cos (2x2 ) + sin (2x2 )) + C
8
4