Tabular integration by parts
Kowalski
Tabular integration by parts. The integration by parts formula can be difficult to apply repeatedly: it takes a lot of space to write down and it’s easy to make a distribution error. Fortunately, there is a purely mechanical procedure for performing integration by parts without writing down so much called tabular integration by parts. It is based on the following theorem and table. High-order integration by parts. Suppose that u0 (x) and v 0 (x) are given functions, and define a sequence of derivatives of u0 by u1 (x) = u0 0 (x),
u2 (x) = u1 0 (x),
...
and define a sequence of antiderivatives of v 0 by Z Z v 1 (x) = v 0 (x) dx, v 2 (x) = v 1 (x) dx,
un (x) = un−1 0 (x) Z
...
v n (x) =
v n−1 (x) dx.
Then the n-th order integration by parts formula is Z Z u0 v 0 dx = u0 v 1 − u1 v 2 + u2 v 3 − u3 v 4 + · · · + (−1)n−1 un−1 v n + (−1)n un v n dx for any integer n > 0. The table. This theorem can be visualized in a table as u u0
dv v0 @
u1
+
@ @ R v1
@
−
@
@ R
u2
v2 @
+
@
@ R
u3
v3 @
−
@
@ R
.. .
v4
.. .
un−2 @ un−1 @ un
±
@ @ R v n−1 ∓
@
@ R
±
R-
vn
In it, each diagonal arrow means to multiply the two functions with the appropriate sign (+ or −), while the lone horizontal arrow at the bottom means to integrate the product of the last two functions (with the appropriate sign).
Tabular IBP algorithm. This technique is often used when the integrand is a product of two different classes of functions. Remember that there are five classes of elementary functions, recalled by the mnemonic LIATE: Logarithms, Inverse trig, Algebraic, Trigonometric, and Exponential.) To use the IBP formula, first choose u and dv. Two helpful (though not mutually exclusive) rules of thumb are: • u: This should be an expression that “simplifies” upon differentiation. Often, the best choice is to set u equal to the first expression present from the term LIATE. • dv: This should be the most “complicated” expression you can integrate, even if you must use a substitution to do so. If there is absolutely nothing you can integrate, set dv = 1 dx. Place u and dv in the top of the appropriate columns and then write each new row in the table by differentiating the u column and integrating the dv column. There are really three main cases to worry about. Case 1. The u column eventually goes to 0, i.e. u = A. If the u column reaches 0, then the antiderivative can be read off the table. • Best bet: This case is well suited for integrals of the form Z £ ¤ £ polynomial · something easy to integrate ] dx. | {z } | {z } u
dv
Z • Example:
x5 sin x dx, with u = x5 and dv = sin(x) dx: u x5
Q
5x4 Q
+
dv sin(x)
Q s Q − cos(x)
−
Q s Q 3 − sin(x) 20x Q + Q s Q 2 cos(x) 60x Q − Q s Q sin(x) 120x Q + Q s Q − cos(x) 120 Q − Q s Q R- − sin(x) 0 +
Hence, the answer is Z x5 sin(x) dx = (x5 )(− cos(x)) − (5x4 )(− sin(x)) + (20x3 )(cos(x)) − (60x2 )(sin(x)) Z + (120x)(− cos(x)) − (120)(− sin(x)) + 0 dx = −x5 cos(x) + 5x4 sin(x) + 20x3 cos(x) − 60x2 sin(x) − 120x cos(x) + 120 sin(x) + C = (5x4 − 60x2 + 120) sin(x) + (−x5 + 20x3 − 120x) cos(x) + C
Case 2. The u column eventually becomes algebraic, i.e. u = L or u = I. Recall that the algebraic functions are those without trigonometric, exponential, logarithmic, or inverse trigonometric terms, so essentially they’re combinations of power functions. In this case: • In the u column, cross out the last entry and write 1 beneath it. • In the dv column, underneath the last entry write the product of the last u and dv terms. Simply do one round of IBP, which should eliminate the logarithm or inverse trig term, and focus on the new integral (which frequently will require a substitution to evaluate). • Best bet: This technique works very well on integrals of the form Z Z £ £ ¤ £ ¤ £ logarithm · algebraic ] dx or inverse trig · algebraic ] dx. | {z } | {z } | {z } | {z } u
u
dv
dv
Z • Example:
x2 ln x dx, with u = ln(x) and dv = x2 dx: u ln(x)
1 x
dv x2 @+ R @ 1 3 Rx − 3
Decoding the table gives Z
µ
¶ Z µ ¶µ ¶ 1 3 1 3 1 x ln(x) dx = (ln(x)) x − x dx 3 3 x Z 1 1 = x3 ln(x) + x2 dx 3 3 1 1 1 = x3 ln(x) + · x3 + C 3 3 3 1 3¡ = x 3 ln(x) + 1) + C 9 2
This is more or less equivalent to doing standard integration by parts, but it omits all the u and dv notation.
Case 3. Both the u and dv columns are periodic (ever repeating), i.e. u = T or u = E. In this case, 1. Apply two rounds of IBP and stop; the entry in the u column should match your original choice of u up to a multiple. 2. Extract the integral equation from the table. 3. Use basic algebra to solve for the unknown integral: move it to the LHS and divide out the multiple in front of the integral. 4. If it’s an indefinite integral, don’t forget the “+C.” This is the most complicated application of IBP, but it’s easy to get the hang of with a little practice. • Best bet: This technique works very well on integrals of the form Z Z £ £ ¤ £ ¤ £ sine/cosine · exponential ] dx or sine/cosine · sine/cosine ] dx. {z } | | | | {z } {z } {z } dv
u
u
dv
Z • Example:
e3x cos 4x dx, with u = cos(4x) and dv = e3x dx: u cos(4x)
dv e3x Q
+
Q
Q Q s
−4 sin(4x) Q
1 3x e 3
Q− Q
−16 cos(4x)
Q Q s 1 3x Re + 9
Hence, the equation we obtain upon the first repeat is µ ¶ µ ¶ Z µ ¶ Z 1 1 1 e3x cos(4x) dx = (cos(4x)) e3x − (−4 sin(4x)) e3x + (−16 cos(4x)) e3x dx 3 9 9 Z 1 3x 4 3x 16 = e cos(4x) + e sin(4x) − e3x cos(4x) dx. 3 9 9 R 3x Now, adding 16 e cos(4x) dx to both sides, we find 9 Z 25 1 4 x3x cos(4x) dx = e3x cos(4x) + e3x sin(4x) 9 3 9 whence Z
µ ¶ 25 1 3x 2 e cos(4x) + e3x sin(4x) 9 3 9 3 3x 4 3x = e cos(4x) + e sin(4x) + C 25 25 ¢ 1 3x ¡ e 3 cos(4x) + 4 sin(4x) + C. = 25
e3x cos(4x) dx =
where the final “+C” is added since this is a general antiderivative.
