Integration by Parts: Formula, Derivation, ILATE Rule & Examples

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Integration by Parts is a method of integration that is used to integrate the product of two or more functions. It is used to find the integrals through the integration of the product of the functions.

Integration by Parts was proposed by Brook Taylor in 1715.
It is also called Partial Integration or Product Rule of Integration.
This method changes the integration of product of functions into integrals for which the solution can be computed easily.
It is used for inverse trigonometric functions and logarithmic functions that do not have integral formulas.

Integration by Parts Formula is given as

∫u v dx = u∫v dx −∫u' (∫v dx) dx

Table of Content

What is Integration by Parts?
Integration By Parts Formula
Integration By Parts Formula Derivation
ILATE Rule
Integration by UV Formula
Integration by Parts with Limits
Applications of Integration by Parts
Integration by Parts Related Formulas
Things to Remember
Previous Years’ Questions
Sample Questions

Key Terms: Integration by Parts, ILATE Rule, Integration by Parts, Partial Integration, Integration by Substitution, Functions

What is Integration by Parts?

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Integration by Parts is an integration method used for integrating the product of two functions. It is also referred to as the Product Rule of Integration.

The two functions f(x) and g(x) that are to be integrated are of form ∫f(x).g(x).
The first function f(x) is chosen such that its derivative formula exists.
The second function g(x) is selected such that its integral exists.

Integration (First Function x Second Function) = (First Function) x (Integration of Second Function) - Integration (Differentiation of First Function x Integration of Second Function).

∫f(x).g(x).dx = f(x) ∫ g(x).dx − ∫ (f′(x) ∫ g(x).dx).dx + C

Integrals Detailed Video Explanation

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Relevant Concepts
Integration of Trigonometric Functions	Double Integral	Continuous Integration
Line Integral	Integral Test	Integrals of Particular Function

Integration By Parts Formula

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Integration by Parts is used to find the integral of the product of two different types of functions. It is used to find integrals for functions such as logarithmic, inverse trigonometric, algebraic, trigonometric, and exponential functions.

Integration By Parts Formula is

∫u v dx = u∫v dx −∫u' (∫v dx) dx

Where

u is the function of u(x).
v is the function v(x).
u' is the derivative of the function u(x).

Integration by Parts Formula

Integration by Parts Formula

Integration by Parts involves the following steps:

Choose u and v
Differentiate u: u'
Integrate v: ∫v dx
Place u, u' and ∫v dx into: u∫v dx −∫u' (∫v dx) dx
Simplify and solve the problem.

By Parts Method of integrals Video Explanation

Solved Example

Example: Determine the integral of x sin2x, using Integration by Parts Formula.

Solution: In order to integrate the given expression, the Integration by Parts Formula needs to be used.

∫ uv.dx = u∫ v.dx -∫( u' ∫ v.dx).dx

Here

u = x
v = Sin2x

∫x sin2x. dx

=x∫sin2xdx - d/dx. x.∫ sin2xdx. dx

=x. -cos2x/2 - ∫(1.-cos2x/2). dx

=-cos2x/2. dx + 1/2 cos2xdx

=-xcos2x/2 + sin2x/4 + C

Therefore, ∫x sin2x dx = -x cos2x/2 +sin 2x/4+ C.

Integration By Parts Formula Derivation

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Integration By Parts Formula can be derived from the formula of the derivative of the product of two functions. For any two given functions f(x) and g(x), the derivative of the product of these two functions is equal to the sum of the derivatives of the first functions multiplied by the second function, and the derivative of the second function multiplied by the first function.

Integration By Parts Formula can be derived using the product rule of differentiation. Assume two functions u and v and let their product be y. i.e., y = uv. Using the product rule of differentiation, we get

d/dx (uv) = u (dv/dx) + v (du/dx)

Rearranging the terms, we get

u (dv/dx) = d/dx (uv) - v (du/dx)

Integrating on both sides with respect to x,

∫ u (dv/dx) (dx) = ∫ d/dx (uv) dx - ∫ v (du/dx) dx

Canceling the terms, we get

∫ u dv = uv - ∫ v du

Thus, Integration By Parts Formula is derived.

ILATE Rule

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ILATE Rule is used in the process of integration by parts for the selection process of the first function and the second function.

What does ILATE mean?

I stands for Inverse Trigonometric Functions
L means Logarithmic Functions
A means Algebraic Functions
T means Trigonometric Functions
E means Exponential Functions

For the integral of the two functions, the left term will be considered the first function while the second term will be considered the second function.

For instance, in x.ex, x will be considered as the first function and ex as the second function.
Thus, the main function will be selected in such a manner that the derivative of the function could be easily integrated.
The preference order of this rule will be dependent on the ILATE rule.

Read More: Integration Rules

Solved Example

Example: Find ∫log x dx.

Solution: Assume that log x is the first function and constant 1 is the second function.

Then the integral of the second function is x. Thus:

→ ∫(logx.1) dx = log x ∫1dx – ∫[d/dx(log x) ∫1 dx]dx

= (log x).x – ∫1/x.x dx

= x log x – x + C

Integration by UV Formula

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Integration by Parts UV Formula is:

∫du (dv/dx) dx = uv−∫v (du/dx) dx

Where

u = Function of u(x)
v = Function of v(x)
dv = Derivative of v(x)
du = Derivative of u(x)

Integration by Parts with Limits

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In calculus, definite integrals are defined as integrals with the upper and lower limits. It is also possible to get the formula of integration by parts with limits. Thus, the formula for the same would be:

∫_a^bdu (dv/ dx)dx = [uv]_a^b−∫_a^bv (du/ dx)dx

Here,

a = Lower Limit
b = Upper Limit

Applications of Integration by Parts

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Integration by Parts is used for the functions or expressions for which the basic integration formulas do not exist and we need to derive the integral. There are no integral solutions for logarithmic functions and for inverse trigonometric functions. Here is the integration of log x and tan^-1x.

Integration of Logarithmic Function

∫ logx.dx = ∫ logx.1.dx

= logx. ∫1.dx - ∫ ((logx)'.∫ 1.dx).dx

=logx.x -∫ (1/x .x).dx

=xlogx - ∫ 1.dx

=x logx - x + C

Integration of Inverse Trigonometric Function

∫ tan^-1x.dx = ∫tan^-1x.1.dx

= tan^-1x.∫1.dx - ∫((tan^-1x)'.∫ 1.dx).dx

= tan^-1x. x - ∫(1/(1 + x²).x).dx

= x. tan^-1x - ∫ 2x/(2(1 + x²)).dx

= x. tan^-1x - ½.log(1 + x²) + C

Integration by Parts Related Formulas

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Given below are the formulas that have been derived from the integration by parts formula. These formulas are used in the process of integration of various algebraic expressions.

∫ e^x(f(x) + f'(x)).dx = e^xf(x) + C
∫√(x² + a²).dx = ½ . x.√(x² + a²)+ a²/2. log|x + √(x² + a²)| C
∫√(x² - a²).dx =½ . x.√(x² - a²) - a²/2. log|x +√(x² - a²) | C
∫√(a² - x²).dx = ½ . x.√(a² - x²) + a²/2. sin^-1x/a + C

Also Check:

Related Topics
List of Integral Formulas	Integration by Partial Fractions	Fundamental Theorem of Calculus
Integral Calculus	Differentiation and Integration Formula	Calculus Formula
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Things to Remember

Integration by Parts is an integration technique used to find the integration of the product of two functions.
It is also referred to as Partial Integration or Product Rule of Integration.
It changes the integration of the product of functions into integrals for which a solution can be derived easily.
Integration by Parts Formula is given as ∫u v dx = u∫v dx −∫u' (∫v dx) dx.
It is used for Inverse Trigonometric Functions and Logarithmic Functions that do not have integral formulas.
ILATE Rule is used for the selection of the first and second function in integration by parts.

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Previous Years’ Questions

\(\int \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} \)is equal to... (BITSAT - 2011)
Let [.] denote the greatest integer function then the value of… (AIEEE - 2011)
\(\displaystyle\int_{1/2}^{2}|\log_{10}\,x|dx=\)...
Evaluate: \(\int\frac{sin x}{sin 4x} dx\)...
Evaluate: \(\int\frac{x^{3}+x}{x^{4} 9}dx\)…
If a is a positive integer, then the number of values...
The integral ∫cos (logx) dx is equal to… (JEE Main - 2019)
The number of integral terms in the expansion…
\(\int\limits_{0} ^{1}\frac{dx}{[ax+b(1 x)]^2}\) is equal to…
Prove that the value of the integral... (JEE Advanced - 1988)

Sample Questions

Ques. What will be the integral of x²e^x? Determine the same using Integration by Parts. (3 Marks)

Ans. According to the ILATE Rule,

u = x²
dv = e^x dx

Then,

du = 2x dx
v = ∫ e^x dx = e^x

Using Integration by Parts Formula,

∫ u dv = uv - ∫ v du

∫ x² e^x dx = x² e^x - ∫ e^x (2x) dx

= x² e^x - 2 ∫ x e^x dx

Using Integration by Parts Formula again to evaluate ∫ x e^x dx,

∫ x² e^x dx = x² e^x - 2 (x e^x - ∫ e^x dx) = x² e^x - 2 x e^x + 2 e^x + C

= e^x (x²- 2 x + 2)+ C

Thus, ∫ x² e^x dx = = e^x (x²- 2 x + 2)+ C

Ques. How to Perform Integration by Parts? (3 Marks)

Ans. To perform Integration by Parts, take f as the first function and g as the second function. Then,

Integration of (First Function x Second Function) = (First Function) x (Integration of Second Function) - Integration of (Differentiation of First Function x Integration of Second Function)

∫f(x).g(x).dx=f(x)∫g(x).dx−∫(f′(x)∫g(x).dx).dx+C

Ques. Explain the Product Rule of Integration. (3 Marks)

Ans. Product Rule of Integration is also referred to as Integration by Parts. It is used to integrate the product of two or more functions. The two functions that have to be integrated, i.e. f(x) and g(x) are of form ∫∫f(x).g(x). Thus, Integration by Parts is also called the product rule of integration. Among the two functions, the first function f(x) is selected in such a manner that its derivative formula exists, and the second function g(x) is chosen such that an integral of such a function exists.

∫f(x).g(x).dx=f(x)∫g(x).dx−∫(f′(x)∫g(x).dx).dx+C

Ques. Can Integration by Parts be used for any Integral? (2 Marks)

Ans. Yes, Integration by Parts can be used for any integral in the method of integrating a function. However, in most cases, it is used for functions or expressions for which the formulas of integration do not exist. It is generally used for:

Logarithmic Functions
Inverse Trigonometric Functions

Ques. List the basic Integration Formulas. (2 Marks)

Ans. Some of the most commonly used Integration Formulas include:

∫ 1 dx = x + C
∫ a dx = ax+ C
∫ xⁿdx = ((xⁿ⁺¹)/(n+1))+C ; n≠1
∫ sin x dx = – cos x + C
∫ cos x dx = sin x + C
∫ sec²x dx = tan x + C
∫ csc²x dx = -cot x + C
∫ sec x (tan x) dx = sec x + C
∫ csc x ( cot x) dx = – csc x + C
∫ (1/x) dx = ln |x| + C
∫ e^xdx = e^x+ C
∫ a^xdx = (a^x/ln a) + C ; a>0, a≠1

Ques. When should I use Integration by Parts? (1 Mark)

Ans. Integration by Parts is used when the simple process of integration is not possible for the given functions. If there are two functions and a product between them is given, integration by parts formula can be used.

Ques. What is the Rule for Integration by Parts? (2 Marks)

Ans. In integration by parts, the product of two functions is first mentioned, and then the required formula is applied. The integral of the two functions is taken, by referring to the left term as the main function and the second term as the second function. This method of selection of the first and second functions is called the ILATE Rule.

Ques. Find: ∫ sec³ x dx. (3 Marks)

Ans. Let I = ∫ sec³ x dx

∫ (sec² x. sec x) dx

secx ∫ sec² x dx – ∫ ( sec x. tan x . tan x) dx

sec x ∫ sec² x dx – ∫ (sec x . tan² x ) dx

sec x. tan x – ∫ ( sec x (sec2 x – 1)) dx

sec x . tan x – ∫ sec3 x dx + ∫ sec x dx

2I = ∫ sec x. tan x + ∫ sec x dx

2I = ∫ sec x. tan x – log | sec x + tan x| + c

I = ½ ( sec x. tan x + log |sec x + tan x| ) + c

Ques. Determine the integral ∫ x ln x dx using Integration by Parts. (3 Marks)

Ans. Using ILATE Rule,

u = ln x
v = x

Using Integration by Parts Formula,

∫ uv dx = u ∫ v dx - ∫ (u' ∫ v dx) dx

∫ x ln x dx = ln x ∫ x dx - ∫ (1/x) (∫ x dx) dx

= ln x (x²/2) - ∫ (1/x) (x²/2) dx

= (x² ln x)/2 - (1/2) ∫ x dx

= (x² ln x) / 2 - (1/2) (x²/2) + C

= (x² ln x) /2 - (x² / 4) + C

=(x²/4)(2 ln x -1) + C

Ques. How to use Limits in Integration by Parts? (2 Marks)

Ans. Limits for Integrations by Parts are used similarly to the definite integrals. On applying the lower limit 'a', and the upper limit 'b' to integration by parts, we get

\(\int^b_a uv.dx =[ u\int v.dx - \int( u'\int v.dx).dx]^b_a\)

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Integration by Parts: Formula, Derivation, ILATE Rule & Examples

What is Integration by Parts?

Integration By Parts Formula

Solved Example

Integration By Parts Formula Derivation

ILATE Rule

Solved Example

Integration by UV Formula

Integration by Parts with Limits

Applications of Integration by Parts

Integration of Logarithmic Function

Integration of Inverse Trigonometric Function

Integration by Parts Related Formulas

Things to Remember

Previous Years’ Questions

Sample Questions

CBSE CLASS XII Related Questions

1.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.
If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

3.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

4.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

6.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

Similar Mathematics Concepts

Comments

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Integration by Parts: Formula, Derivation, ILATE Rule & Examples

What is Integration by Parts?

Integration By Parts Formula

Solved Example

Integration By Parts Formula Derivation

ILATE Rule

Solved Example

Integration by UV Formula

Integration by Parts with Limits

Applications of Integration by Parts

Integration of Logarithmic Function

Integration of Inverse Trigonometric Function

Integration by Parts Related Formulas

Things to Remember

Previous Years’ Questions

Sample Questions

CBSE CLASS XII Related Questions

1.Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

3.Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

4.A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

6. Find: The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.
If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

3.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

4.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

6.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is: