LibraryIntegration by Parts

Integration by Parts

Learn about Integration by Parts as part of JEE Mathematics Mastery - Calculus and Algebra

Mastering Integration by Parts for Competitive Exams

Integration by Parts is a fundamental technique in calculus, crucial for solving integrals that cannot be solved by simple substitution. It's a cornerstone for many problems encountered in competitive exams like JEE Mathematics.

The Core Formula

The integration by parts formula is derived from the product rule of differentiation. If we have two functions, u(x)u(x) and v(x)v(x), their product's derivative is given by: ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}. Integrating both sides with respect to xx, we get: uv=udvdxdx+vdudxdxuv = \int u\frac{dv}{dx} dx + \int v\frac{du}{dx} dx. Rearranging this, we arrive at the integration by parts formula:

The Integration by Parts Formula: $\int u dv = uv - \int v du$

This formula allows us to transform a difficult integral into a potentially simpler one by choosing appropriate parts for uu and dvdv.

The formula is udv=uvvdu\int u dv = uv - \int v du. The key to using this formula effectively lies in choosing the functions uu and dvdv wisely. The goal is to make the new integral, vdu\int v du, simpler to solve than the original integral, udv\int u dv.

Choosing 'u' and 'dv': The LIATE Rule

A common mnemonic to help choose which function to assign as uu (and consequently dvdv) is LIATE. This acronym stands for:

LIATE
LogarithmicInverse TrigonometricAlgebraicTrigonometricExponential

The function that appears earliest in the LIATE order should generally be chosen as uu. This is because differentiating logarithmic and inverse trigonometric functions often simplifies them, while integrating algebraic, trigonometric, and exponential functions usually results in expressions of similar complexity or slightly more complex forms. Therefore, making the function that simplifies upon differentiation our uu is usually the best strategy.

Remember: The goal is to make vdu\int v du easier to solve than udv\int u dv. LIATE is a guideline, not a strict rule; sometimes, you might need to deviate based on the specific integral.

Illustrative Example

Let's solve the integral xcos(x)dx\int x \cos(x) dx using integration by parts.

To solve xcos(x)dx\int x \cos(x) dx, we apply the integration by parts formula: udv=uvvdu\int u dv = uv - \int v du. Following the LIATE rule, we choose u=xu = x (Algebraic) and dv=cos(x)dxdv = \cos(x) dx (Trigonometric). From these choices, we find du=dxdu = dx and v=cos(x)dx=sin(x)v = \int \cos(x) dx = \sin(x). Substituting these into the formula gives: xcos(x)dx=xsin(x)sin(x)dx\int x \cos(x) dx = x \sin(x) - \int \sin(x) dx. The remaining integral sin(x)dx\int \sin(x) dx is straightforward to solve, yielding cos(x)-\cos(x). Therefore, the final result is xsin(x)(cos(x))+Cx \sin(x) - (-\cos(x)) + C, which simplifies to xsin(x)+cos(x)+Cx \sin(x) + \cos(x) + C.

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Handling Repeated Application

Some integrals require applying the integration by parts formula multiple times. This often happens when the product of an algebraic function and a trigonometric or exponential function is involved, and the algebraic part is of a higher degree (e.g., x2exx^2 e^x). In such cases, each application reduces the power of the algebraic term until it becomes a constant.

Special Cases and Tricks

There are specific types of integrals where integration by parts is particularly useful, such as integrals involving exsin(x)e^x \sin(x) or excos(x)e^x \cos(x). These often lead to an equation where the original integral appears on both sides, allowing you to solve for it algebraically. Also, be mindful of integrals where you might need to choose dvdv such that it's not immediately obvious how to integrate it, or where uu might be a constant.

What is the primary goal when choosing 'u' and 'dv' in integration by parts?

To make the new integral vdu\int v du simpler to solve than the original integral udv\int u dv.

Which function type typically comes first in the LIATE mnemonic?

Logarithmic functions.

Learning Resources

Khan Academy: Integration by Parts(video)

A clear and concise video explanation of the integration by parts formula and its application with examples.

Paul's Online Math Notes: Integration by Parts(documentation)

Comprehensive notes covering the formula, LIATE rule, and various examples, including repeated applications.

Brilliant.org: Integration by Parts(blog)

An interactive explanation of integration by parts with engaging examples and conceptual insights.

MIT OpenCourseware: Calculus - Integration by Parts(documentation)

Problem set solutions from MIT that often include detailed walkthroughs of integration by parts problems.

YouTube: Integration by Parts - JEE Mathematics(video)

A video specifically tailored for competitive exams like JEE, focusing on common problem types and strategies for integration by parts.

MathWorld: Integration by Parts(documentation)

A detailed mathematical treatment of the integration by parts formula, including its derivation and properties.

Symbolab Blog: How to Use Integration by Parts(blog)

A practical guide with step-by-step instructions and examples on applying the integration by parts technique.

Coursera: Calculus Specialization - Integration Techniques(tutorial)

A structured course module that covers integration by parts as part of a broader set of integration techniques.

Wikipedia: Integration by Parts(wikipedia)

A comprehensive overview of the integration by parts formula, its history, and applications in various fields of mathematics.

Art of Problem Solving: Integration by Parts(documentation)

A resource from a community focused on challenging math problems, offering insights and advanced techniques for integration by parts.