Integrating Logarithmic and Trigonometric Functions: A Comprehensive Guide

Welcome to this detailed guide on integrating logarithmic and trigonometric functions. This article aims to provide a thorough understanding of how to integrate complex expressions, specifically focusing on the integration of the expression int; (log x)/sqrt{1 - x^2} dx. We will explore the techniques of integration by parts and partial fractions to find the solution step-by-step.

Introduction

Integration by parts is a powerful method used to integrate products of functions. It is based on the product rule for differentiation and is particularly useful when dealing with expressions involving logarithmic and trigonometric functions. By breaking down the problem into smaller, more manageable parts, we can tackle even the most complex integrals.

Integration by Parts: A Step-by-Step Approach

Let's consider the integral int; (log x)/sqrt{1 - x^2} dx. The first step is to identify suitable functions for u and dv. We set:

u log x, which gives du 1/x dx

dv 1/sqrt{1 - x^2} dx, which implies v arcsin(x)

Applying the integration by parts formula, we get:

int; u dv uv - int; v du

Substituting our choices, we have:

int; (log x)/sqrt{1 - x^2} dx log x cdot arcsin x - int; arcsin x cdot (1/x) dx

Next, we need to evaluate the integral int; arcsin x cdot (1/x) dx. This requires another application of integration by parts. We set:

u arcsin x, which gives du (1/sqrt{1-x^2}) dx

dv (1/x) dx, which gives v log x

Thus, we have:

int; arcsin x cdot (1/x) dx arcsin x cdot log x - int; log x cdot (1/sqrt{1-x^2}) dx

Notice that the integral int; log x cdot (1/sqrt{1-x^2}) dx is the same as our original integral. Let's denote the original integral as I:

I int; (log x)/sqrt{1 - x^2} dx

Therefore, we can express it as:

I log x cdot arcsin x - arcsin x cdot log x - I

This simplifies to:

0 0

indicating that we need to sum the parts correctly:

2I log x cdot arcsin x

Thus, the final result for the integral is:

int; (log x)/sqrt{1 - x^2} dx (1/2) log x cdot arcsin x C

Evaluating the Integral Using Integration by Parts and Partial Fractions

Let's now evaluate the integral:

int; log left(x/sqrt{1-x^2}right) dx

Using integration by parts, we set:

u log left(x/sqrt{1-x^2}right), which gives du (1/x) dx

dv dx, which implies v x

Thus, we have:

int; u dv uv - int; v du

Substituting our choices, we get:

x log left(x/sqrt{1-x^2}right) - int; x cdot (1/x) cdot (sqrt{1-x^2}/x) cdot (sqrt{1-x^2} - x cdot (-2x)/(2 sqrt{1-x^2}))/(1-x^2) dx

Simplifying the integrand yields:

int; log left(x/sqrt{1-x^2}right) dx x log left(x/sqrt{1-x^2}right) - int; (1/(1-x^2)) dx

Further simplification gives us:

int; (1/(1-x^2)) dx (1/2) log left((1-x)/(1 x)right)

Therefore, the final result is:

int; log left(x/sqrt{1-x^2}right) dx x log left(x/sqrt{1-x^2}right) - (1/2) log left((1-x)/(1 x)right) C

Conclusion

In this guide, we have explored the integration of logarithmic and trigonometric functions using the techniques of integration by parts and partial fractions. We have provided detailed steps and examples to help you understand and apply these methods effectively. Whether you are a math student or a professional working with complex integrals, this comprehensive guide will undoubtedly be a valuable resource.