Integration by parts is a technique used in calculus to find the integral of the product of two functions. It’s essentially a reversal of the product rule for differentiation.
Integrating a function is not always easy sometimes we have to integrate a function that is the multiple of two or more functions in this case if we have to find the integration we have to use integration by part concept, which uses two products of two functions and tells us how to find their integration.
Now let’s learn about Integration by parts, its formula, derivation, and others in detail in this article.
Table of Content
What is Integration by Parts?
Integration by part is the technique used to find the integration of the product of two or more functions where the integration can not be performed using normal techniques. Suppose we have two functions f(x) and g(x) and we have to find the integration of their product i.e., âˆ« f(x).g(x) dx where it is not possible to further solve the product of this product f(x).g(x).Â
This integration is achieved using the formula:
âˆ« f(x).g(x) dx = f(x) âˆ« g(x) d(x) – âˆ« [f'(x) {âˆ«g(x) dx} dx] dx + c
where f'(x) is the first differentiation of f(x).
This formula is read as:
Integration of the First Function multiplied by the Second Function is equal to (First Function) multiplied by (Integration of Second Function) – Integration of (Differentiation of First Function multiplied by Integration of Second Function).
From the above formula, we can easily observe that choosing the first function and the second function is very important for the success of this formula, and how we choose the first function and the second function is discussed further in this article.
What is Partial Integration?
Partial integration, also known as integration by parts, is a technique used in calculus to evaluate the integral of a product of two functions. The formula for partial integration is given by:
âˆ« u dv = uv – âˆ« v du
where u and v are differentiable functions of x. This formula allows us to simplify the integral of a product by breaking it down into two simpler integrals. The idea is to choose u and dv in such a way that the new integral on the righthand side is easier to evaluate than the original integral on the lefthand side. This technique is particularly useful when dealing with products of functions that do not have simple antiderivatives.
History of Partial Integration
Integration by part concept was first proposed by the famous Brook Taylor in his book in 1715. He wrote that we can find the integration of the product of two functions whose differentiation formulas exist. Some important functions do not have integrations formulas and their integration is achieved using integration by parttaking them as a product of two functions. For example, âˆ«ln x dx can not be calculated using normal integration techniques. But we can integrate it using Integration by part technique and taking it as a product of two functions that is, Â âˆ«1.ln x dx.
Integration By Parts Formula
Integration by parts formula is the formula that helps us to achieve the integration of the product of two or more functions. Suppose we have to integrate the product of two functions as
âˆ«u.v dxÂ
where u and v are the functions of x, then this can be achieved using,
âˆ«u.v dx = u âˆ« v d(x) – âˆ« [u’ {âˆ«v dx} dx] dx + c
The order to choose the First function and the Second function is very important and the concept used in most of the cases to find the first function and the second function is ILATE concept.
Using the above formula and the ILATE concept we can easily find the integration of the product of two functions. The integration by part formula is shown in the image below,
ILATE Rule
The ILATE rule tells us about how to choose the first function and the second function while solving the integration of the product of two functions. Suppose we have two functions of x u and v and we have to find the integration of their product then we choose the first function and the by ILATE rule.
The ILATE full form is discussed in the image below,
The ILATE rules give us the hierarchy of taking the first function, i.e. if in the given product of the function, one function is a Logarithmic function and another function is a Trigonometric function. Now we take the Logarithmic function as the first function as it comes above in the hierarchy of the ILATE rule similarly, we choose the first and second functions accordingly.
NOTE: It is not always appropriate to use the ILATE rule sometimes other rules are also used to find the first function and the second function.
Integration By Parts Formula Derivation
Integration By Parts Formula is derived using the product rule of differentiation. Suppose we have two functions u and v and x then the derivative of their product is achieved using the formula,
d/dx (uv) = u (dv/dx) + v (du/dx)
Now to derive the integration by parts formula using the product rule of differentiation.
Rearranging the terms
u (dv/dx) = d/dx (uv) – v (du/dx)
Integrating both sides with respect to x,
âˆ« u (dv/dx) (dx) = âˆ« d/dx (uv) dx – âˆ« v (du/dx) dx
simplifying,
âˆ« u dv = uv – âˆ« v du
Thus, the integration by parts formula is derived.
How to Â Find Integration by Part?
Integration by part is used to find the integration of the product of two functions. We can achieve this using the steps discussed below,
Suppose we have to simplify âˆ«uv dx
Step 1: Choose the first and the second function according to the ILATE rule. Suppose we take u as the first function and v as the second function.
Step 2: Differentiate u(x) with respect to x that is, Evaluate du/dx.
Step 3: Integrate v(x) with respect to x that is, Evaluate âˆ«v dx.
Use the results obtained in Step 1 and Step 2 in the formula,
âˆ«uv dx = uâˆ«v dx âˆ’ âˆ«((du/dx)âˆ«v dx) dx
Step 4: Simplify the above formula to get the required integration.
Repeated Integration by Parts
Repeated integration by parts is an extension of the integration by parts technique in calculus. It is used when you have a product of functions that requires integration multiple times to find the antiderivative. The process involves applying the integration by parts formula iteratively until you reach a point where the resulting integral is easy to evaluate or has a known form.
When applying this formula repeatedly, you would start with an integral that involves a product of two functions, and then apply integration by parts to break it down into simpler integrals. You would then continue this process on the resulting integrals until you reach a point where further applications are unnecessary or where the integrals become manageable.
Here’s a stepbystep example of how repeated integration by parts works:
 Start with an integral of a product of two functions: âˆ« u dv.
 Apply the integration by parts formula to get: uv – âˆ« v du.
 If the new integral obtained on the righthand side still involves a product of functions, apply integration by parts again to break it down further.
 Continue this process until you obtain a simpler integral that can be easily evaluated or one that matches a known integral form.
Tabular Integration by Parts
Tabular integration, also known as the tabular method or the method of tabular integration, is an alternative technique for evaluating integrals that involve repeated application of integration by parts. This method is particularly useful when dealing with integrals where the product of functions can be integrated multiple times to reach a simple result.
The tabular method organizes the repeated integration by parts process into a table, making it easier to keep track of the terms and simplify the integral efficiently. Here’s how the tabular method works:
 Begin by writing down the functions involved in the integral in two columns: one for the function to differentiate (u) and another for the function to integrate (dv).
 Start with the function to integrate (dv) on the left column and the function to differentiate (u) on the right column.
 Continue differentiating the function in the u column until you reach zero or a constant. At each step, integrate the function in the dv column until you reach a point where further integration is not necessary.
 Multiply the terms diagonally and alternate the signs (+ and ) for each term. Sum up these products to find the result of the integration.
Here’s an example to illustrate the tabular integration method:
Let’s evaluate the integral âˆ«x sin(x) dx.
 Step 1: Create a table with two columns for u (function to differentiate) and dv (function to integrate):
u  dv 

x  sin(x) 
 Step 2: Differentiate the function in the u column and integrate the function in the dv column:
u  dv 

x  cos(x) 
1  sin(x) 
0  cos(x) 
 Step 3: Multiply the terms diagonally and alternate the signs:
(x)(cos(x)) – (1)(sin(x)) + (0)(cos(x)) = x cos(x) + sin(x)
So, the result of the integral âˆ«x sin(x) dx is xcos(x) + sin(x).
The tabular integration method is especially useful when dealing with integrals that involve functions that repeat upon differentiation or integration, allowing for a systematic and organized approach to finding the antiderivative.
Applications of Integration by Parts
Integration by Parts has various applications in integral calculus it is used to find the integration of the function where normal integration techniques fail. We can easily find the integration of inverse and logarithmic functions using the integration by parts concept.
We will find the Integration of the Logarithmic function and Arctan function using integration by part rule,
Integration of Logarithmic Function (log x)
Integration of Inverse Logarithmic Function (log x) is achieved using Integration by part formula. The integration is discussed below,
âˆ« logx.dx = âˆ« logx.1.dx
Taking log x as the first function and 1 as the second function.
Using âˆ«u.v dx = u âˆ« v d(x) – âˆ« [u’ {âˆ«v dx} dx] dx
â‡’ âˆ« logx.1.dx = Â logx. âˆ«1.dx – âˆ« ((logx)’.âˆ« 1.dx).dx
â‡’ âˆ« logx.1.dx = logx.x âˆ« (1/x .x).dx
â‡’ âˆ« logx.1.dx = xlogx – âˆ« 1.dx
â‡’ âˆ« logx.dx = x logx – x + C
Which is the required integration of logarithmic function.
Integration of Inverse Trigonometric Function (tan^{1} x)
Integration of Inverse Trigonometric Function (tan^{1} x) is achieved using Integration by part formula. The integration is discussed below,
âˆ« tan^{1}x.dx = âˆ«tan^{1}x.1.dx
Taking tan^{1} x as the first function and 1 as the second function.
Using âˆ«u.v dx = u âˆ« v d(x) – âˆ« [u’ {âˆ«v dx} dx] dx
â‡’ âˆ«tan^{1}x.1.dx = tan^{1}x.âˆ«1.dx – âˆ«((tan^{1}x)’.âˆ« 1.dx).dx
â‡’ âˆ«tan^{1}x.1.dx = tan^{1}x. x – âˆ«(1/(1 + x^{2}).x).dx
â‡’ âˆ«tan^{1}x.1.dx = x. tan^{1}x – âˆ« 2x/(2(1 + x^{2})).dx
â‡’ âˆ«tan^{1}x.dx = x. tan^{1}x – Â½.log(1 + x2) + C
Which is the required integration of Inverse Trigonometric Function.
Reallife Applications of Partial Integration
Some of the common real life application of partial integration are:
 Finding Antiderivatives
 In engineering and physics, partial integration is used to find antiderivatives of functions that represent physical quantities. For example, in mechanics, it’s used to derive equations of motion from the equations of force and acceleration.
 Wallis Product
 The Wallis product, an infinite product representation of pi, can be derived using partial integration techniques. This product has applications in fields such as number theory, probability theory, and signal processing.
 Gamma Function Identity
 The gamma function, which extends the factorial function to complex numbers, has various applications in mathematics, physics, and engineering. Partial integration is used to prove identities involving the gamma function, which are crucial in areas like probability theory, statistical mechanics, and quantum mechanics.
 Use in Harmonic Analysis
 Partial integration plays a significant role in harmonic analysis, particularly in Fourier analysis. It is used to derive properties of Fourier transforms, such as the convolution theorem and properties of Fourier series. These results are applied in fields like signal processing, image analysis, and telecommunications.
Formulas of Integration by Parts
We can derive the integration of various functions using the integration by parts concept. Some of the important formulas derived using this technique are
 âˆ« e^{x}(f(x) + f'(x)).dx = e^{x}f(x) + C
 âˆ«âˆš(x^{2} + a^{2}).dx = Â½ . x.âˆš(x^{2} + a^{2})+ a^{2}/2. logx + âˆš(x^{2} + a^{2}) + C
 âˆ«âˆš(x^{2} – a^{2}).dx =Â½ . x.âˆš(x^{2} – a^{2}) – a^{2}/2. logx +âˆš(x^{2} – a^{2})  Â C
 âˆ«âˆš(a^{2} – x^{2}).dx = Â½ . x.âˆš(a^{2} – x^{2}) + a^{2}/2. sin^{1} x/a + C
Integration By Parts Examples
Example 1: Find âˆ« e^{x }x dx.
Solution:
Let I = Â âˆ« e^{x} x dx
Choosing u and v using ILATE rule
u = x
v = e^{x}Differentiating u
u'(x) = d(u)/dx
â‡’ u'(x) = d(x)/dx
â‡’ u'(x) = 1
âˆ«v dx = âˆ«e^{x} dx = e^{x}
Using the Integration by part formula,
â‡’ I = âˆ« e^{x} x dxÂ
â‡’ I = x âˆ«e^{x }dx âˆ’ âˆ«1 (âˆ« e^{x }dx) dx
â‡’ I = xe^{x }âˆ’ e^{x }+ C
â‡’ I = e^{x}(x âˆ’ 1) + C
ExampleÂ 2: Calculate âˆ« x sin x dx.
Solution:
Let I = âˆ« x sin x dx
Choosing u and v using ILATE rule
u = x
v = sin xDifferentiating u
u'(x) = d(u)/dx
â‡’ u'(x) = d(x)/dx
â‡’ u'(x) = 1
Using the Integration by part formula,
â‡’ I = âˆ« x sin x dxÂ
â‡’ I = x âˆ«sin x dx âˆ’ âˆ«1 âˆ«(sin x dx) dx
â‡’ I = âˆ’ x cos x âˆ’ âˆ«âˆ’cos x dx
â‡’ I = âˆ’ x cos x + sin x + C
Example 3: Find âˆ« sin^{âˆ’1} x Â dx.
Solution:
Let I= Â âˆ« sin^{âˆ’1} x Â dx
â‡’ I = Â âˆ« 1.sin^{âˆ’1} x Â dx
Choosing u and v using ILATE rule
u = sin^{âˆ’1} x
v = 1Differentiating u
u'(x) = d(u)/dx
â‡’ u'(x) = d(sin^{âˆ’1} x )/dxÂ
â‡’ u'(x) = 1/âˆš(1 âˆ’ x ^{2})Â
Using the Integration by part formula,
â‡’ I = âˆ« sin^{âˆ’1} x dxÂ
â‡’ I = sin^{âˆ’1} x âˆ« 1 dx Â âˆ’ âˆ« 1/âˆš(1 âˆ’ x ^{2}) Â âˆ«(1 dx) Â dx
â‡’ I Â = x sin^{âˆ’1 }x Â âˆ’ âˆ«( x/âˆš(1 âˆ’ x ^{2} ) )dx
Let, t = 1 âˆ’ x ^{2}Â
Differentiating both sides
Â dt = âˆ’2x dx
â‡’ âˆ’dt/2 = x dx
â‡’ I = âˆ« sin^{âˆ’1} x dx = Â x sin^{âˆ’1} x Â âˆ’ âˆ«âˆ’(1/2âˆšt ) dt
â‡’ I Â = x sin^{âˆ’1 }x Â + 1/2âˆ«t^{âˆ’1/2 }dt
â‡’ I Â = x sin^{âˆ’1} x + t^{1/2} + C
â‡’ I Â = x sin^{âˆ’1} x + âˆš(1 âˆ’ x^{2 }) + C
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FAQs on Integration by Parts
What is integration by parts?
Integration by parts is the technique for finding the integration of the product of the two functions where the normal techniques of integration fail. Integration by the part formula is the,
âˆ«u.v dx = u âˆ« v d(x) – âˆ« [u’ {âˆ«v dx} dx] dx + c
What is integration by parts formula?
For two functions f(x) and g(x) the integration by part formula is,
âˆ« f(x).g(x) dx = f(x) âˆ« g(x) d(x) – âˆ« [f'(x) {âˆ«g(x) dx} dx] dx + c
where f'(x) is differentiation of f(x).
How to derive integration by parts formula?
Integration by part formula is derived using the product rule of differentiation.
Why do we use integration by parts formula?
Integration by part formula is used to find the integration of the function when the normal differentiation techniques fail. We can find the integration of inverse trigonometric functions, and logarithmic functions using Integration by part formula
What is the application of integration by parts?
Integration by part has various applications and the basic application of it is that it is used to find the integration of the function when the function is given as the product of the functions which can not be simplified further. For example âˆ« f(x).g(x) dx is achieved using Integration by parts.