Figure 1.
Part of a student's work citing the LIATE rule to conduct integration by parts.
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Integration by parts is a powerful technique for obtaining solutions for many types of integration commonly involved in differential equations [1,2,3,4,5], integral transforms and other advanced mathematics [6,7,8], hence it is a necessary part included in all calculus textbooks for most undergraduate STEM programs in the world [9,10,11,12,13]. Integration by parts is originated from the inverse operation of the product rule of differentiation expressed as follows
| $ \int {udv} = uv - \int {vdu} , $ | (1) |
where both u and v can be different types of functions. In practice, once u and dv on the left side of formula (1) are identified, v and du can be obtained through
| $ \int {udv} = uv - \int {vdu} \xleftarrow{{}}v = \int {dv} , \ \ \ \ \ du = u'dx . $ | (2) |
This formula seems expressed simply but choosing the order for u and v to make the integral $ \int {vdu} $ simpler than $ \int {udv} $ depends more on individual's experience.
In teaching applied calculus courses to undergraduate engineering and education students for more than a decade at a regional university in Australia, through answering numerous questions from different cohorts of students on how to better decide the order of u and v in using integration by parts, the author also observed the practices of students who adopted different approaches to solve the same questions at different times. For example, several years ago a few students cited the 'LIATE rule' for choosing the order of u and v in using integration by parts to solve the following indefinite integral
| $ \int {(9{x^2}\ln x - x\cos x)dx} , $ |
as shown in Figure 1. Their attempts were successful in finding the integral for the question, which led the author to find Kasube's classic note on LIATE published in 1983 [14] where L refers to the logarithmic functions; I refers to the inverse trigonometric functions; A refers to the algebraic functions; T refers to the trigonometric functions, and E refers to the exponential functions.
Further investigations revealed that LIATE is more a rough 'rule of thumb' to guide choosing the order of u and v in integration by parts for some types of combined integrands, far from a universal rule considered by many students as their default guide in using integration by parts. Such misunderstanding caused more confusions for the students who tried to apply LIATE to some seemingly simple integrals but actually without exact solutions. For example, at the time when students were learning how to use numeric methods to obtain approximate results for $ \int\limits_0^1 {\frac{{\sin x}}{x}dx} $, some students tried to use LIATE to obtain an exact solution for the integral. Even after they were told this seemingly simple integral has no exact solution, they believed that the integrand as a combination of A and T was perfect for applying LIATE.
Nevertheless, LIATE seems a useful guide for students to decide the order in integration by parts for many common combinations of functions given the fact no other similar rules are available. Hence, from that time onwards, LIATE was introduced to the students as a rough rule to guide their practices of using integration by parts whereas the early draft of this technical note was also prepared to balance the potential over-reliance of LIATE to which students may tend. With the accumulation of more incompatible cases with LIATE in teaching integrations to undergraduate students in the recent years, this technical note has been gradually enriched to the current status ready to be shared with the mathematics teaching and learning community in the world.
In Section 2, the LIATE rule is reviewed using examples to not only highlight its strengths in guiding the use of integration by parts but also reveal its weaknesses or limits. Section 3 presents the Pet-LoPo-InPo guide for using integration by parts with examples. Concluding remarks are summarized in Section 4. It should be noted that the new guide reported in this note only intends to provide students with a practical guideline that can reduce potential confusions caused by the incompatibility of some types of combinative integrands indicated in LIATE. It is not a universal rule to cover all possible integrable integrands through integration by parts.
In 1983, Kasube [14] proposed LIATE as a rule of thumb for choosing the u function that comes first in $ \int {udv} = uv - \int {vdu} $. LIATE stands for
● L – logarithmic functions, for example, lnx, logbx
● I – inverse trigonometric functions, such as arcsinx, arctanx
● A – algebraic functions, for example, 2x3, 3x, –5
● T – trigonometric functions, for example, sinx, cos2x, 4tan3x
● E – exponential functions, for example, ex, e–3x, 52x.
LIATE indicates the order of selecting the first function (first part) u from the combined integrand by integration by parts. Once u is selected, the remainder of the combined term become the differential dv.
For example, in $ \int {x{e^{ - x}}dx} $, the combined integrand is $ x{e^{ - x}} $. By LIATE, x is classified as A and should be chosen as u ahead of e–x (E), or u = x and $ dv = {e^{ - x}}dx $.
Example 1: Evaluate $ \int {x{e^{ - x}}dx} $.
In the integrand $ x{e^{ - x}} $, x is an algebraic function (A) and e–x is an exponential function (E). By formula (2) and LIATE, let $ u = x{\text{ and }}dv = {e^{ - x}}dx $.
|
$ ∫x_ue−xdx_dv=(x_u)(−e−x_v)−∫(−e−x_v)(dx_du)←v=∫dv=∫e−xdx=−e−x, du=u′dx=(x)′dx=dx =−xe−x+∫e−xdx=−xe−x+(−e−x)+c=−xe−x−e−x+c=−e−x(x+1)+c. $
|
Example 2: Evaluate $ \int {{x^3}\ln xdx} $.
In the integrand $ {x^3}{e^{ - x}} $, x3 is an algebraic function (A) and lnx is a logarithmic function (L). By formula (2) and LIATE, let $ u = \ln x{\text{ and }}dv = {x^3}dx $.
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$ ∫lnx_ux3dx_dv=(lnx_u)(14x4_v)−∫(14x4_v)(dxx_du)←v=∫x3dx=14x4, du=(lnx)′dx=dxx =14x4lnx−14∫x3dx=14x4lnx−116x4+c=116x4(4lnx−1)+c. $
|
Example 3: Evaluate $ \int {2x\cos xdx} $.
In the integrand $ 2x\cos x $, 2x is an algebraic function (A) and cosx is a trigonometric function (T). By formula (2) and LIATE, let $ u = 2x{\text{ and }}dv = \cos xdx $.
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$ ∫2x_ucosxdx_dv=(2x_u)(sinx_v)−∫(sinx_v)(2dx_du)←v=∫cosxdx=sinx, du=(2x)′dx=2dx =2xsinx−∫2sinxdx=2xsinx−2(−cosx)+c=2xsinx+2cosx+c. $
|
Example 4: Evaluate $ \int {2\arcsin xdx} $.
In the integrand $ 2\arcsin x $, 2 is a constant (A) and arcsinx is an inverse trigonometric function (I). By formula (2) and LIATE, let $ u = \arcsin x{\text{ and }}dv = 2dx $.
|
$ ∫arcsinx_u(2dx)_dv=(arcsinx_u)(2x_v)−∫(2x_v)(dx√1−x2_du)←v=2x, du=(arcsinx)′dx=dx√1−x2 =2xarcsinx−∫2xdx√1−x2←t=1−x2,dt=−2xdx↔2xdx=−dt =2xarcsinx−∫−dt√t=2xarcsinx+2√t+c=2xarcsinx+2√1−x2+c. $
|
Example 5: Evaluate $ \int {{x^2}\arctan xdx} $.
In the integrand $ {x^2}\arctan x $, x2 is an algebraic function (A) and arctanx is an inverse trigonometric function (I). By formula (2) and LIATE, let $ u = \arctan x{\text{ and }}dv = {x^2}dx $.
| $ \int {\mathop {\underline {\arctan x} }\limits_u \mathop {\underline {({x^2}dx)} }\limits_{dv} } = (\mathop {\underline {\arctan x} }\limits_u )(\mathop {\underline {\frac{1}{3}{x^3}} }\limits_v ) - \int {\left( {\mathop {\underline {\frac{1}{3}{x^3}} }\limits_v } \right)\left( {\mathop {\underline {\frac{{dx}}{{1 + {x^2}}}} }\limits_{du} } \right)} \xleftarrow{{}}v = \frac{1}{3}{x^3}, {\text{ }}du = (\arctan x)'dx = \frac{{dx}}{{1 + {x^2}}} $ |
|
$ =13x3 arctan x−13∫x31+x2dx←x31+x2=x−x1+x2 by long division =13x3 arctan x−13∫(x−x1+x2)dx=13x3 arctan x−13(12x2−∫x1+x2dx) =13x3 arctan x−16x2+13∫x1+x2dx←u=1+x2,du=2xdx↔xdx=12du =13x3 arctan x−16x2+13∫12duu=13x3 arctan x−16x2+16lnu+c =16[2x3 arctan x−x2+ln(1+x2)]+c. $
|
Although LIATE works well in finding out the integrals for some types of combinative integrands, more types of combinative integrands that are within the span of LIATE cannot be integrated using integration by parts under the guidance of LIATE. For example, the combinations of LI (e.g., $ \ln x \cdot \arctan x $), LT (e.g., $ \ln x \cdot \cos x $), LE (e.g., $ \ln x \cdot {e^x} $), IT (e.g., $ \arcsin x \cdot \cos x $), IE (e.g., $ \arctan x \cdot {e^x} $), and some AT (e.g., $ \frac{{\sin x}}{x} $) are indicated by LIATE, but none of these is integrable by integration by parts
Example 6: Evaluate $ \int {\ln x\cos xdx} $.
In the integrand $ \ln x\cos x $, lnx is a logarithmic function (L) and cosx is a trigonometric function (T). By formula (2) and LIATE, let $ u = \ln x{\text{ and }}dv = \cos xdx $.
| $ \int {\mathop {\underline {\ln x} }\limits_u \mathop {\underline {\cos xdx} }\limits_{dv} } = (\mathop {\underline {\ln x} }\limits_u )(\mathop {\underline {\sin x} }\limits_v ) - \int {(\mathop {\underline {\sin x} }\limits_v )\left( {\mathop {\underline {\frac{1}{x}dx} }\limits_{du} } \right)} = \ln x\sin x - \int {\frac{{\sin x}}{x}dx} . $ |
Here $ \int {\frac{{\sin x}}{x}dx} $ is not integrable by analytical methods.
Example 7: Evaluate $ \int {\arctan x \cdot {e^x}dx} $.
In the integrand arctanx·ex, $ \arctan x $ is an inverse trigonometric function (I) and ex is an exponential function (E). By formula (2) and LIATE, let u = arctanx and dv = exdx.
| $ \int {\mathop {\underline {\arctan x} }\limits_u \mathop {\underline {{e^x}dx} }\limits_{dv} } = (\mathop {\underline {\arctan x} }\limits_u )(\mathop {\underline {{e^x}} }\limits_v ) - \int {(\mathop {\underline {{e^x}} }\limits_v )\left( {\mathop {\underline {\frac{{dx}}{{1 + {x^2}}}} }\limits_{du} } \right)} = {e^x}\arctan x - \int {\frac{{{e^x}}}{{1 + {x^2}}}dx} . $ |
Here $ \int {\frac{{{e^x}}}{{1 + {x^2}}}dx} $ is not integrable by analytical methods.
Hence, without an explicit explanation on the limit or applicability of LIATE, students might be misled to some extent.
Another ambiguous issue with LIATE is that its term for algebraic functions (A) in LIATE is too loose compared with other functions that are specified for particular operations. For example, logarithmic functions are defined for any positive number to any positive base except 1. In mathematics, algebraic functions cover all common functions, including power functions, polynomials, and fractions. In most cases, A in LIATE only represents polynomial functions where the exponent only takes zero or positive integers. This issue has been noticed by some educators and one suggestion was to replace A in LIATE by P for polynomial functions, which makes LIATE to LIPTE [15]. By using P for A, however, LIPTE would exclude some combinative integrands involving power functions with negative or fractional values for the exponent, which is demonstrated by the following examples.
Example 8: Evaluate $ \int {\frac{1}{{\sqrt x }}\ln xdx} $.
In the integrand $ \frac{1}{{\sqrt x }}\ln x $, lnx is a logarithmic function (L) and $ \frac{1}{{\sqrt x }} = {x^{ - \frac{1}{2}}} $ is an algebraic function (A). By formula (2) and LIATE, let $ u = \ln x{\text{ and }}dv = {x^{ - \frac{1}{2}}}dx $.
|
$ ∫lnx_u1√xdx_dv=(lnx_u)(2√x_v)−∫(2√x_v)(1xdx_du)=2√xlnx−∫2√xdx =2√xlnx−4√x+c=2√x(lnx−2)+c. $
|
However, by LIPTE, $ \frac{1}{{\sqrt x }} = {x^{ - \frac{1}{2}}} $ is not a polynomial function; hence LIPTE is not applicable to it.
Example 9: Evaluate $ \int {\frac{1}{{{x^3}}}\ln xdx} $.
In the integrand $ \frac{1}{{{x^3}}}\ln x $, lnx is a logarithmic function (L) and $ \frac{1}{{{x^3}}} = {x^{ - 3}} $ is an algebraic function (A). By formula (2) and LIATE, let $ u = \ln x{\text{ and }}dv = \frac{1}{{{x^3}}}dx $.
| $ \int {\mathop {\underline {\ln x} }\limits_u \mathop {\underline {\frac{1}{{{x^3}}}dx} }\limits_{dv} } = (\mathop {\underline {\ln x} }\limits_u )\left( {\mathop {\underline { - \frac{1}{{2{x^2}}}} }\limits_v } \right) - \int {\left( {\mathop {\underline { - \frac{1}{{2{x^2}}}} }\limits_v } \right)\left( {\mathop {\underline {\frac{1}{x}dx} }\limits_{du} } \right)} = - \frac{1}{{2{x^2}}}\ln x + \int {\frac{1}{{2{x^3}}}dx} = - \frac{1}{{2{x^2}}}\ln x - \frac{1}{{4{x^2}}} + c \\ \ \ \ \ \ = - \frac{1}{{4{x^2}}}(2\ln x + 1) + c. \\ $ |
However, by LIPTE, $ \frac{1}{{{x^3}}} = {x^{ - 3}} $ is not a polynomial function; hence LIPTE is not applicable to this case.
Another minor issue with LIATE is that the order for a combinative integrand with a trigonometric function (T) and an exponential function (E) is interchangeable in most cases if it is integrable. Hence, TE can be as ET too, which is demonstrated by the following example.
Example 10: Evaluate $ \int {{e^x}\cos xdx} $.
In the integrand excosx, cosx is a trigonometric function (T) and ex is an exponential function (E). By formula (2) and LIATE, let $ u = \cos x{\text{ and }}dv = {e^x}dx $.
| $ I = \int {\mathop {\underline {\cos x} }\limits_u \mathop {\underline {{e^x}dx} }\limits_{dv} } = (\mathop {\underline {\cos x} }\limits_u )(\mathop {\underline {{e^x}} }\limits_v ) - \int {(\mathop {\underline {{e^x}} }\limits_v )(\mathop {\underline { - \sin xdx} }\limits_{du} )} = {e^x}\cos x + \int {\sin x{e^x}dx} \\ \ \ \ \ \ = {e^x}\cos x + \left( {\sin x{e^x} - \int {\cos x{e^x}dx} } \right) = {e^x}\cos x + {e^x}\sin x - \mathop {\underline {\int {\cos x{e^x}dx} } }\limits_I \\ I = {e^x}\sin x + {e^x}\cos x - I\xrightarrow{{}}2I = {e^x}\sin x + {e^x}\cos x \\ I = \int {\cos x{e^x}dx} = \frac{1}{2}{e^x}(\sin x + \cos x) + c. \\ $ |
Alternatively, the same result can be found by exchanging the order of the exponential function ex (E) with the trigonometric function cosx (T) as follows.
| $ I = \int {\mathop {\underline {{e^x}} }\limits_u \mathop {\underline {\cos xdx} }\limits_{dv} } = (\mathop {\underline {{e^x}} }\limits_u )(\mathop {\underline {\sin x} }\limits_v ) - \int {(\mathop {\underline {\sin x} }\limits_v )(\mathop {\underline {{e^x}dx} }\limits_{du} )} = {e^x}\sin x - \int {\sin x{e^x}dx} \\ \ \ \ \ \ = {e^x}\sin x - \left( {{e^x}( - \cos x) + \int {\cos x{e^x}dx} } \right) = {e^x}\sin x + {e^x}\cos x - \mathop {\underline {\int {{e^x}\cos xdx} } }\limits_I \\ I = {e^x}\sin x + {e^x}\cos x - I\xrightarrow{{}}2I = {e^x}(\sin x + \cos x) \\ I = \int {\cos x{e^x}dx} = \frac{1}{2}{e^x}(\sin x + \cos x) + c. \\ $ |
This kind of technique is sometimes referred to as the embedded integration by parts [13,16]. Even though, in most cases, only sine or cosine functions combined with either exponential functions or polynomial functions are integrable by the embedded integration by parts, rather than all trigonometric functions, which is demonstrated by the following example.
Example 11: Evaluate $ \int {{x^2}\tan xdx} $.
In the integrand x2tanx, tanx is a trigonometric function (T) and x2 is a polynomial function (P). By formula (2) and LIPTE, let $ u = {x^2}{\text{ and }}dv = \tan xdx $.
| $ \int {\mathop {\underline {{x^2}} }\limits_u \mathop {\underline {\tan xdx} }\limits_{dv} } = (\mathop {\underline {{x^2}} }\limits_u )(\mathop {\underline { - \ln \cos x} }\limits_v ) - \int {(\mathop {\underline { - \ln \cos x} }\limits_v )(\mathop {\underline {2xdx} }\limits_{du} )} = - {x^2}\ln \cos x + \int {2x\ln \cos xdx} . $ |
The new integration has an integrand $ 2x\ln \cos x $, a combination of a logarithmic function lncosx and a polynomial function 2x, which by LIPTE should go with $ u = \ln \cos x{\text{ and }}dv = 2xdx $. This will result in
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$ ∫x2_utanxdx_dv=−x2lncosx+∫2xlncosxdx=−x2lncosx+(lncosx)(x2)−∫x2(−tanxdx)or ∫x2_utanxdx_dv=−x2lncosx+x2lncosx+∫x2tanxdx=∫x2tanxdx. $
|
This means such integration is not resolvable by integration by parts, despite indicated in LIPTE.
By examining LIATE and LIPTE, the algebraic function (A) or the polynomial function (P) plays a pivotal role in both of them. As aforementioned, either A or P is not properly defined in LIATE or LIPTE and simply using A or P would not be a proper reflection of the integrability of the combined integrands and the function types of A or P. Hence, a new guide based on the properties of the P-functions, Pet-LoPo-InPo, is summarized in this note for conducting integration by parts. The Pet-LoPo-InPo guide is based on the following classified definitions.
● P: Polynomial functions in the form of axn where n can take zero and positive integers, for example, 5, 2x2, –4x5, (3x2 – 4x5 + 1).
● Po: Power functions in the form of axn where n can take any real number (zero, positive and negative), for example, 3, 2x3, x–2, $ \sqrt x $, $ \frac{2}{{\sqrt[3]{x}}} $. In most cases, n cannot take –1.
● nPo: Integer power functions in the form of axn where n can only take integers, for example, 3, 2x3, x–2. In most cases, n cannot take –1.
● Lo: Logarithmic functions, for example, lnx, logbx.
● I: Inverse trigonometric functions, such as arcsinx, arctanx.
● e: Exponential functions, for example, ex, e–3x, 52x.
● t: The primary trigonometric (or sinusoidal) functions, for example, sinx, cos2x, 4sin3x.
The three sub-guides, Pet, LoPo and InPo, are independent from each other, i.e., each sub-guide only works within its group. Pet means that combinations of any two functions from polynomial functions, exponential functions, and primary trigonometric functions are possibly integrable by integration by parts. LoPo indicates that combinations of logarithmic functions only with power functions are possibly integrable by integration by parts. Similarly, InPo implies that combinations of inverse trigonometric functions only with integer power functions are possibly integrable by integration by parts. This is a stark contrast to LIATE that does not exclude those incompatible combinations for using integration by parts.
Within a sub-guide, the process works the same as using LIATE, i.e., the order of selecting the first function (first part) u from the combined integrand for integration by parts is from the left letter to the right letter. For example, in the integrand 3x2cosx, 3x2 is a polynomial function (P) and cosx is a primary trigonometric function (t). By the Pet guide, P (3x2) is on the left of t (cosx), hence, u = 3x2.
In Pet, capital P means that any polynomial function should come first before either the exponential or primary trigonometric functions in integration by parts. The use of small letters e and t implies that the order of these two functions is interchangeable in integration by parts. This is to relax the order indicated in LIATE. The use of Pet-LoPo-InPo is demonstrated by the following examples.
The Pet guide is largely similar to ATE in LIATE or PTE in LIPTE demonstrated in Examples 1, 3 and 10, except the interchangeability of the order between the exponential and primary trigonometric functions in their combinations.
Example 12: Evaluate $ \int {(2{x^2} - 4x + 3)\sin 2xdx} $.
In $ (2{x^2} - 4x + 3)\sin 2x $, $ (2{x^2} - 4x + 3) $ is a second-order polynomial function (P) and sin2x is a primary trigonometric function (t). By Pet, let $ u = 2{x^2} - 4x + 3 $ and $ dv = \sin 2xdx $.
| $ \int {\mathop {\underline {(2{x^2} - 4x + 3)} }\limits_u \mathop {\underline {\sin 2xdx} }\limits_{dv} } = \mathop {\underline {(2{x^2} - 4x + 3)} }\limits_u \left( {\mathop {\underline { - \frac{1}{2}\cos 2x} }\limits_v } \right) - \int {\left( {\mathop {\underline { - \frac{1}{2}\cos 2x} }\limits_v } \right)\mathop {\underline {(4x - 4)dx} }\limits_{du} } \\ \ \ \ \ \ = - \left( {{x^2} - 2x + \frac{3}{2}} \right)\cos 2x + \int {2(x - 1)\cos 2xdx} \\ \ \ \ \ \ = - \left( {{x^2} - 2x + \frac{3}{2}} \right)\cos 2x + 2(x - 1)\left( {\frac{1}{2}\sin 2x} \right) - \int {\sin 2xdx} \\ \ \ \ \ \ = \left( { - {x^2} + 2x - \frac{3}{2}} \right)\cos 2x + (x - 1)\sin 2x + \frac{1}{2}\cos 2x + c \\ \ \ \ \ \ = ( - {x^2} + 2x - 1)\cos 2x + (x - 1)\sin 2x + c = (x - 1)\sin 2x - {(x - 1)^2}\cos 2x + c. \\ $ |
Example 13: Evaluate $ \int {(3{x^2} - 4x)({e^x} - {e^{ - x}})dx} $.
In $ (3{x^2} - 4x)({e^x} - {e^{ - x}}) $, $ (3{x^2} - 4x) $ is a second-order polynomial function (P) and $ ({e^x} - {e^{ - x}}) $ is a combined exponential function (e). By Pet, let $ u = 3{x^2} - 4x $ and $ dv = ({e^x} - {e^{ - x}})dx $.
| $ \int {\mathop {\underline {(3{x^2} - 4x)} }\limits_u \mathop {\underline {({e^x} - {e^{ - x}})dx} }\limits_{dv} } = \mathop {\underline {(3{x^2} - 4x)} }\limits_u \mathop {(\underline {{e^x} + {e^{ - x}})} }\limits_v - \int {\mathop {(\underline {{e^x} + {e^{ - x}})} }\limits_v \mathop {\underline {(6x - 4)dx} }\limits_{du} } \\ \ \ \ \ \ = (3{x^2} - 4x)({e^x} + {e^{ - x}}) - \int {\mathop {\underline {(6x - 4)} }\limits_u \mathop {(\underline {{e^x} + {e^{ - x}})dx} }\limits_{dv} } \\ \ \ \ \ \ = (3{x^2} - 4x)({e^x} + {e^{ - x}}) - (6x - 4)({e^x} - {e^{ - x}}) + \int {6({e^x} - {e^{ - x}})dx} \\ \ \ \ \ \ = (3{x^2} - 4x)({e^x} + {e^{ - x}}) - (6x - 4)({e^x} - {e^{ - x}}) + 6({e^x} + {e^{ - x}}) + c \\ \ \ \ \ \ = (3{x^2} - 4x + 6)({e^x} + {e^{ - x}}) - (6x - 4)({e^x} - {e^{ - x}}) + c \\ \ \ \ \ \ = (3{x^2} - 10x + 10){e^x} + (3{x^2} + 2x + 2){e^{ - x}} + c \\ $ |
The interchangeability of the order between the exponential and primary trigonometric functions in their combinations was already demonstrated in Example 10.
The LoPo guide is for integrands only involving both logarithmic functions (Lo) and power functions (Po) because logarithmic functions are not compatible with any other functions except power functions for using integration by parts. In such a situation, the logarithmic function should always come first before the power function. The LoPo guide not only excludes the unnecessary relationships between logarithmic functions and other functions in LIATE but also broadens the coverage of the P-functions in LIPTE. Examples 8 and 9 solved previously already demonstrated the power of LoPo for such cases. The following three examples demonstrate both the special and general cases of using LoPo for integration by parts, respectively.
Example 14: Evaluate $ I = \int {\frac{{\ln x}}{x}dx} $.
As the integrand involves lnx, by LoPo, let $ u = \ln x $ and $ dv = \frac{1}{x}dx $.
| $ I = \int {\mathop {\underline {\ln x} }\limits_u \mathop {\underline {\frac{1}{x}dx} }\limits_{dv} } = \mathop {\underline {\ln x} }\limits_u \mathop {\underline {\ln x} }\limits_v - \int {\mathop {\underline {\ln x} }\limits_v } \left( {\mathop {\underline {\frac{{dx}}{x}} }\limits_{du} } \right) = {(\ln x)^2} - \int {\frac{{\ln x}}{x}dx} = {(\ln x)^2} - I $ |
| $ 2I = {(\ln x)^2}\xrightarrow{{}}I = \frac{1}{2}{(\ln x)^2} + c. $ |
Alternatively, this integral can be resolved by substitution but it is not the focus of this note.
Example 15: Evaluate $ I(n) = \int {{x^n}\ln xdx} $ where n is any real number except –1.
In $ {x^n}\ln x $, $ {x^n} $ is a power function (Po). By LoPo, let $ u = \ln x $ and $ dv = {x^n}dx $.
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$ I(n)=∫lnx_uxndx_dv=lnx_u(1n+1xn+1)_v−∫(1n+1xn+1)_v(dxx)_du=1n+1xn+1lnx−1n+1∫xndx =1n+1xn+1lnx−1(n+1)2xn+1+c=1n+1xn+1(lnx−1n+1)+c. $
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This is a general solution to all integrands in the form of $ {x^n}\ln x $ where n can take any real number, except –1. For example, taking n = 3, the solution is the same as that in Example 2; taking n = –3, the solution is the same as that in Example 9; taking n = –1/2, the solution is the same as that in Example 8. By taking n = 1/3, the solution becomes
| $ I(\frac{1}{3}) = \int {{x^{\frac{1}{3}}}\ln xdx} = \frac{1}{{\frac{1}{3} + 1}}{x^{\frac{1}{3} + 1}}\left( {\ln x - \frac{1}{{\frac{1}{3} + 1}}} \right) + c = \frac{3}{4}{x^{\frac{4}{3}}}\left( {\ln x - \frac{3}{4}} \right) + c . $ |
This formula is not applicable for n = –1, which is the special case solved in Example 14.
The following example is to derive a reduction formula using integration by parts.
Example 16: Evaluate $ I(m, n) = \int {{{(\ln x)}^m}{x^n}dx} $ where m is any positive integer and n is any real number except –1.
The integrand is a combination of the logarithmic function (Lo) and the power function (Po). By LoPo, let $ u = {(\ln x)^m} $ and $ dv = {x^n}dx $.
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$ I(m,n)=∫(lnx)m_uxndx_dv=(lnx)m_u(1n+1xn+1)_v−∫(1n+1xn+1)_v(m(lnx)m−1dxx)_du=1n+1(lnx)mxn+1−mn+1∫(lnx)m−1xndx=1n+1(lnx)mxn+1−mn+1I(m−1,n). $
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This is actually the general reduction formula to all integrands in the form of $ {(\ln x)^m}{x^n} $ where m is any positive integer and n can take any real number except –1. This means that Examples 2, 8, 9 and 15 are special cases of I(1, 3), I(1, –½), I(1, –3), and I(1, n), respectively. To obtain the integral for I(2, 3), we can use the above reduction formula as follows.
| $ I(2, 3) = \int {{{(\ln x)}^2}{x^3}dx} = \frac{1}{{3 + 1}}{(\ln x)^2}{x^{3 + 1}} - \frac{2}{{3 + 1}}I(2 - 1, 3) = \frac{1}{4}{(\ln x)^2}{x^4} - \frac{1}{2}I(1, 3) . $ |
I(1, 3) was obtained in Example 2, hence,
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$ I(2,3)=∫(lnx)2x3dx=14(lnx)2x4−12I(1,3)=14(lnx)2x4−12[116x4(4lnx−1)]+c=14(lnx)2x4−132x4(4lnx−1)+c=132x4[8(lnx)2−4lnx+1]+c. $
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Similarly, the integral for $ {(\ln x)^3} $ can be found using this reduction formula as flows.
| $ I(0, 0) = \int {{{(\ln x)}^0}{x^0}dx} = \int {dx = x} \\ I(1, 0) = \int {(\ln x)dx} = \frac{1}{{0 + 1}}(\ln x){x^{0 + 1}} - \frac{1}{{0 + 1}}I(1 - 1, 0) = (\ln x)x - I(0, 0) = (\ln x)x - x \\ I(2, 0) = \int {{{(\ln x)}^2}dx} = \frac{1}{{0 + 1}}{(\ln x)^2}{x^{0 + 1}} - \frac{2}{{0 + 1}}I(2 - 1, 0) = {(\ln x)^2}x - 2I(1, 0) \\ = {(\ln x)^2}x - 2\left[ {(\ln x)x - x} \right] = x\left[ {{{(\ln x)}^2} - 2(\ln x) + 2} \right] \\ I(3, 0) = \int {{{(\ln x)}^3}dx} = \frac{1}{{0 + 1}}{(\ln x)^3}{x^{0 + 1}} - \frac{3}{{0 + 1}}I(3 - 1, 0) = {(\ln x)^3}x - 3I(2, 0) \\ = {(\ln x)^3}x - 3x\left[ {{{(\ln x)}^2} - 2(\ln x) + 2} \right] + c = x\left[ {{{(\ln x)}^3} - 3{{(\ln x)}^2} + 6(\ln x) - 6} \right] + c. $ |
InPo is for integrands involving some combinations of the inverse trigonometric functions (I) and the integer power functions (nPo) where the exponent can only take negative or positive integers, except –1. In such a situation, the inverse trigonometric function should always come first before the power function. The InPo guide not only excludes the unnecessary relationships between the inverse trigonometric functions and other functions in LIATE but also expands the coverage of the polynomial functions in LIPTE to the negative half. Examples 4 and 5 solved previously can also be solved exactly the same way using InPo. The following two examples demonstrate solving integrals involving power functions with negative integer exponents by InPo through integration by parts.
Example 17: Evaluate $ \int {\frac{{\arcsin x}}{{{x^3}}}dx} = \int {{x^{ - 3}}\arcsin xdx} $.
The integrand is a combination of an inverse trigonometric function and a power function with a negative index. By InPo, let $ u = \arcsin x $ and $ dv = {x^{ - 3}}dx $.
| $ \int {\mathop {\underline {\arcsin x} }\limits_u \mathop {\underline {{x^{ - 3}}dx} }\limits_{dv} } = \mathop {\underline {(\arcsin x)} }\limits_u \mathop {\underline {\left( {\frac{1}{{ - 2}}{x^{ - 2}}} \right)} }\limits_v - \int {\mathop {\underline {\left( {\frac{1}{{ - 2}}{x^{ - 2}}} \right)} }\limits_v } \mathop {\underline {\left( {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \right)} }\limits_{du} = - \frac{1}{{2{x^2}}}\arcsin x + \frac{1}{2}\int {\frac{{dx}}{{{x^2}\sqrt {1 - {x^2}} }}} . $ |
Since
|
$ ∫dxx2√1−x2=∫costdtsin2t√1−sin2t=∫costdtsin2tcost←x=sint,dx=costdt =∫dtsin2t=−cott=−costsint=−√1−x2x, $
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| $ \int {\frac{{\arcsin x}}{{{x^3}}}dx} = - \frac{1}{{2{x^2}}}\arcsin x - \frac{1}{2}\frac{{\sqrt {1 - {x^2}} }}{x} + c = - \frac{1}{{2{x^2}}}(\arcsin x + x\sqrt {1 - {x^2}} ) + c . $ |
Example 18: Evaluate $ \int {\frac{{\arctan x}}{{{x^4}}}dx} = \int {{x^{ - 4}}\arctan xdx} $.
The integrand is a combination of an inverse trigonometric function and a power function with a negative index. By InPo, let $ u = \arctan x $ and $ dv = {x^{ - 4}}dx $.
| $ \int {\mathop {\underline {\arctan x} }\limits_u \mathop {\underline {{x^{ - 4}}dx} }\limits_{dv} } = \mathop {\underline {(\arctan x)} }\limits_u \mathop {\underline {\left( {\frac{1}{{ - 3}}{x^{ - 3}}} \right)} }\limits_v - \int {\mathop {\underline {\left( {\frac{1}{{ - 3}}{x^{ - 3}}} \right)} }\limits_v } \mathop {\underline {\left( {\frac{{dx}}{{1 + {x^2}}}} \right)} }\limits_{du} = - \frac{1}{{3{x^3}}}\arctan x + \frac{1}{3}\int {\frac{{dx}}{{{x^3}(1 + {x^2})}}} . $ |
By partial fractions
| $ \frac{1}{{{x^3}(1 + {x^2})}} = \frac{{a{x^2} + bx + c}}{{{x^3}}} + \frac{{dx + e}}{{1 + {x^2}}} = \frac{{(a{x^2} + bx + c)(1 + {x^2}) + {x^3}(dx + e)}}{{{x^3}(1 + {x^2})}} $ |
| $ \frac{1}{{{x^3}(1 + {x^2})}} = \frac{{a{x^2} + bx + c + a{x^4} + b{x^3} + c{x^2} + d{x^4} + e{x^3}}}{{{x^3}(1 + {x^2})}} = \frac{{(a + d){x^4} + (b + e){x^3} + (a + c){x^2} + bx + c}}{{{x^3}(1 + {x^2})}} $ |
Hence, by comparing the coefficients on both sides,
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$ \left\{ a+d=0b+e=0a+c=0b=0c=1 \right.\xrightarrow{{}}\left\{ d=−a=1e=−b=0a=−c=−1b=0c=1 \right.\xrightarrow{{}}\frac{1}{{{x^3}(1 + {x^2})}} = \frac{{1 - {x^2}}}{{{x^3}}} + \frac{x}{{1 + {x^2}}} = \frac{1}{{{x^3}}} - \frac{1}{x} + \frac{x}{{1 + {x^2}}} . $
|
Therefore,
| $ \int {\frac{{\arctan x}}{{{x^4}}}dx} = - \frac{1}{{3{x^3}}}\arctan x + \frac{1}{3}\int {\frac{{dx}}{{{x^3}(1 + {x^2})}}} = - \frac{1}{{3{x^3}}}\arctan x + \frac{1}{3}\int {\left( {\frac{1}{{{x^3}}} - \frac{1}{x} + \frac{x}{{1 + {x^2}}}} \right)dx} $ |
| $ = - \frac{1}{{3{x^3}}}\arctan x + \frac{1}{3}\left( { - \frac{1}{{2{x^2}}} - \ln x + \frac{1}{2}\ln (1 + {x^2})} \right) + c \\ = - \frac{1}{{3{x^3}}}\arctan x - \frac{1}{{6{x^2}}} - \frac{1}{3}\ln x + \frac{1}{6}\ln (1 + {x^2}) + c = \frac{1}{6}\left[ {\ln \frac{{1 + {x^2}}}{{{x^2}}} - \frac{1}{{{x^3}}}(2\arctan x + x)} \right] + c. \\ $ |
This note is based on the author's tutorial notes on integration by parts for the undergraduate students accumulated over the past several years at a regional university in Australia. It was triggered by the citation of the LIATE rule by a few students in their assignments prior to the preparation of this note on integration by parts. Although further investigations found some problems with LIATE for conducting integration by parts, LIATE remains a useful rule of thumb for undergraduate students in their attempts to solve integrals through integration by parts. Nevertheless, it is desirable to amend those problems with LIATE so as to reduce the possible causes that may confuse or mislead students in utilizing integration by parts. It would be even better to expand the span of LIATE to cover more types of integrands solvable through integration by parts or by mixed approaches.
By reclassifying the algebraic functions (A) in LIATE into the polynomial functions (P), the standard power functions (Po), and the integer power functions (nPo), this note has largely achieved the goals by modifying LIATE or LIPTE to Pet-LoPo-InPo to guide practices of using integration by parts. The Pet guide applies to integrands involving any combination of two functions from polynomial, exponential, and primary trigonometric functions, indicating that the polynomial function should come first before either the exponential or the primary trigonometric function whereas the order of the exponential and primary trigonometric functions are interchangeable. The LoPo guide applies to the integrands only involving logarithmic and power functions where the logarithmic function is always the first choice. The InPo guide applies to the integrands involving only the inverse trigonometric and the integer power functions where the inverse trigonometric function is always the first choice.
This new guide removes many incompatible combinative integrands included in LIATE, rationalizes the relationship between the exponential and trigonometric functions in LIATE, and expands the coverage of the P-functions beyond the traditional definition in LIPET. Hence, the new guide can reduce confusions experienced by students in using LIATE for their practices of integration by parts. The advantages of this new guide have been demonstrated by many worked examples in this note.
However, this new guide is not as concise as the LIATE rule. Furthermore, although this new guide has expanded the span of LIATE for using integration by parts, it is still unsure how much more the Pet-LoPo-InPo guide could reach out to other integrands by using integration by parts. As demonstrated in the worked examples, particularly for InPo where the inverse trigonometric functions are involved, integration by parts alone would not be able to find the final result. Mixed approaches, demonstrated in Examples 4, 5, 17 and 18, would be required to obtain the integrals for sophisticated integrands.
The author declares there is no conflict of interest.
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| 1. | William Guo, Solving problems involving numerical integration (Ⅱ): Modified Simpson's methods for equal intervals of odd numbers, 2023, 3, 2767-1925, 171, 10.3934/steme.2023011 | |
| 2. | William Guo, Solving problems involving numerical integration (I): Incorporating different techniques, 2023, 3, 2767-1925, 130, 10.3934/steme.2023009 |
Figures(8)
Joe A. Lemire, Marc A. Demeter, Iain George, Howard Ceri, Raymond J. Turner. A novel approach for harnessing biofilm communities in moving bed biofilm reactors for industrial wastewater treatment[J]. AIMS Bioengineering, 2015, 2(4): 387-403. doi: 10.3934/bioeng.2015.4.387
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