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Home > News & Articles > What is ILATE Rule? (With Examples): Derivation, Concepts, Formula, Illustrations
Updated on 18th July, 2023 , 4 min read
Integration is a fundamental concept in calculus that allows us to find the antiderivative of a function. One powerful technique for solving integrals is called integration by substitution, commonly known as ILATE. Integration by substitution is also termed as the full form of ILATE. This method allows us to simplify complicated integrals by substituting variables. By choosing appropriate substitutions and following the rule and formula, we can transform challenging integrals into more manageable forms. Through practice and understanding, mastering ILATE enables us to solve a wide range of integration problems efficiently.
Also Read: UV Rule of Integration
The ILATE Rule In integration, ILATE, can be stated as follows:
∫f(g(x)) * g'(x) dx = ∫f(u) du (ILATE Rule Formula)
(where u = g(x) and du = g'(x) dx. In other words, we substitute the variable u for g(x) and differentiate to find du. Then we rewrite the integral in terms of u and du)
In other words, this rule states that if we have a function f(g(x)) multiplied by the derivative of the inner function g(x), denoted as g'(x), we can rewrite the integral in terms of the new variable u = g(x) and its differential du = g'(x) dx. By doing so, we simplify the integral and make it easier to evaluate.
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The integration by-parts formula can be derived using the following steps:
The integration by parts formula can be used to evaluate integrals of the product of two functions. The first function, u(x), is called the integrand, and the second function, v'(x), is called the integrand's derivative. The integral of the product of two functions is easier to evaluate if the first function is easier to differentiate.
To apply ILATE formula for integration, follow these steps:
1. Identify a function within the integral that can be written as the composition of two functions, such as f(g(x)).
2. Let u be the inner function, g(x), and differentiate it to find du.
3. Rewrite the integral in terms of u and du.
4. Evaluate the new integral in terms of u.
5. If necessary, convert the result back to the original variable x.
Must Read: Differentiation and Integration
Let's look at a few examples to illustrate the application of ILATE/ILATE integration formula:
Evaluate ∫x^2 * (x^3 + 1)^4 dx.
In this example, we can let u = x^3 + 1. Differentiating, we get du = 3x^2 dx. Rearranging, dx = du / (3x^2).
Substituting u and dx in terms of u, the integral becomes:
∫(u - 1)^4 * (du / (3x^2))
Simplifying further:
(1/3)∫(u - 1)^4 / x^2 du
Now, integrate with respect to u:
(1/3) * (u^5/5 - 4u^4/4 + 6u^3/3 - 4u^2/2 + u) + C
Finally, substitute u back in terms of x to get the final answer.
Evaluate ∫x * e^(x^2) dx.
In this example, we can let u = x^2. Differentiating, we get du = 2x dx. Rearranging, dx = du / (2x).
Substituting u and dx in terms of u, the integral becomes:
(1/2)∫e^u du
Now, integrate with respect to u:
(1/2) * e^u + C
Finally, substitute u back in terms of x to obtain the final answer.
Evaluate ∫x^2 * e^(x^3) dx.
In this example, we can let u = x^3. Differentiating, we get du = 3x^2 dx. Rearranging, dx = du / (3x^2).
Substituting u and dx in terms of u, the integral becomes:
(1/3) ∫e^u du
Now, integrate with respect to u:
(1/3) * e^u + C
Finally, substitute u back in terms of x to obtain the final answer.
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Evaluate ∫x * sqrt(x^2 + 1) dx.
In this example, we can let u = x^2 + 1. To find du, we differentiate u with respect to x: du = 2x dx. Rearranging, we have dx = du / (2x).
Substituting u and dx in terms of u, the integral becomes:
(1/2) ∫sqrt(u) du
Now, integrate with respect to u:
(1/2) * (2/3) * u^(3/2) + C
Finally, substitute u back in terms of x to obtain the final answer.
Evaluate ∫x * sin(x^2) dx.
In this example, we can let u = x^2. Differentiating, we get du = 2x dx. Rearranging, dx = du / (2x).
Substituting u and dx in terms of u, the integral becomes:
(1/2) ∫sin(u) du
Now, integrate with respect to u:
-(1/2) * cos(u) + C
Finally, substitute u back in terms of x to obtain the final answer.
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By - Nikita Parmar 2024-09-06 10:59:22 , 6 min readIntegration by parts is a technique used to find the integral of a product of two functions. It is based on the product rule of differentiation and provides a method to transform an integral into a more manageable form.
Integration by parts involves the following formula: ∫u dv = uv - ∫v du. Here, u and v are functions that are chosen based on their differentiability and simplification properties. The goal is to choose u and dv in such a way that the resulting integral on the right side of the formula is easier to solve.
The formula for integration by parts is: ∫u dv = uv - ∫v du. This formula helps in breaking down an integral of a product into two simpler integrals.
When choosing u and dv, it is generally helpful to prioritize functions that become simpler or reduce in complexity when differentiated. Common choices for u include logarithmic functions (ln), inverse trigonometric functions (arcsin, arccos, arctan), algebraic functions (x^n), or exponential functions (e^x). The remaining part of the integrand becomes dv.
Integration by parts is typically used when the integral involves a product of functions and direct integration techniques like substitution or simple rules (such as power rule) are not applicable. It is especially useful when there is a polynomial multiplied by a transcendental function, or when the integral contains products of logarithmic, exponential, or trigonometric functions.
Yes, integration by parts can be applied multiple times to solve more complex integrals. Each application of the technique reduces the complexity of the integral, making it more manageable
Yes, there are alternative integration techniques, such as substitution, trigonometric identities, and partial fraction decomposition. These methods can be effective for specific types of integrals, so it’s always beneficial to explore different approaches when solving integration problems.
One common mistake is choosing u and dv incorrectly. It is essential to select u and dv in a way that simplifies the integral. Another mistake is not applying integration by parts multiple times when necessary, as some integrals may require repeated application of the technique.
Here are some examples of the ILATE rule: ∫ sin(x) ln(x) dx In this example, the logarithmic function comes first in the ILATE order, so we should differentiate the sine function first. ∫ e^x cos(x) dx In this example, the exponential function comes first in the ILATE order, so we should differentiate the cosine function first. ∫ 1/√x tan(x) dx In this example, the inverse trigonometric function comes first in the ILATE order, so we should differentiate the tangent function first.
The full form of ILATE is Inverse, Logarithmic, Algebraic, Trigonometric, Exponential. It is a mnemonic device used to help students choose the first function to differentiate when using integration by parts.
The ILATE rule priority order is: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential.
The acronym ILATE stands for Inverse, Logarithmic, Algebraic, Trigonometric, Exponential.
Integration by parts is a technique for finding the integral of the product of two functions.
The integration by parts rule formula is: ∫ u(x) v'(x) dx = u(x) v(x) − ∫ u'(x) v(x) dx
The product rule of integration is a method for integrating the product of two functions. It is a generalization of the product rule of differentiation, and it can be used to integrate products of any two functions.