Problem 45
Question
Evaluate the indefinite integral as an infinite series. $$\int \frac{\cos x-1}{x} d x$$
Step-by-Step Solution
Verified Answer
The integral is \( \int \frac{\cos x - 1}{x} \, dx = -\frac{x^2}{4} + \frac{x^4}{96} - \frac{x^6}{4320} + \dots + C \).
1Step 1: Express the Function as a Series
We use the Maclaurin series expansion for \( \cos x \):\[\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \dots\]Now, substitute \( \cos x - 1 \) in the integrand:\[\cos x - 1 = - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \dots\]
2Step 2: Divide Each Term by x
Now, divide each term of the series \( \cos x - 1 \) by \( x \):\[\frac{\cos x - 1}{x} = - \frac{x}{2} + \frac{x^3}{24} - \frac{x^5}{720} + \dots\]
3Step 3: Integrate Term by Term
Integrate each term of the series:1. \( \int -\frac{x}{2} \, dx = -\frac{x^2}{4} + C_1 \)2. \( \int \frac{x^3}{24} \, dx = \frac{x^4}{96} + C_2 \)3. \( \int -\frac{x^5}{720} \, dx = -\frac{x^6}{4320} + C_3 \)Continuing, the general form of the integrated term \( \int \frac{x^{2n+1}}{a_n} \, dx \) is:\[\frac{x^{2n+2}}{a_n(2n+2)}\]
4Step 4: Write the Integrated Series
Combine each integrated term to form the series:\[\int \frac{\cos x - 1}{x} \, dx = -\frac{x^2}{4} + \frac{x^4}{96} - \frac{x^6}{4320} + \dots + C\]where \( C \) is the constant of integration.
Key Concepts
Maclaurin SeriesIndefinite IntegralTerm-by-Term Integration
Maclaurin Series
The Maclaurin series is a specific kind of Taylor series centered at zero. Essentially, it is a way to express functions as an infinite sum of terms calculated from derivatives of the function at a single point.
When working with trigonometric, logarithmic or exponential functions, Maclaurin series can be incredibly helpful in mathematics to transform complex functions into simpler polynomial approximations. This is especially useful in calculus when we need to work with functions that are otherwise difficult to integrate or differentiate.
For example, in the case of the exercise at hand, we utilize the Maclaurin series of the function \(\cos x \) which is expressed as:
When working with trigonometric, logarithmic or exponential functions, Maclaurin series can be incredibly helpful in mathematics to transform complex functions into simpler polynomial approximations. This is especially useful in calculus when we need to work with functions that are otherwise difficult to integrate or differentiate.
For example, in the case of the exercise at hand, we utilize the Maclaurin series of the function \(\cos x \) which is expressed as:
- \(\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \dots \)
Indefinite Integral
An indefinite integral, also known as an antiderivative, is the reverse process of differentiation. The main goal is to find a function whose derivative is the given function.
When integrating, we seek a function \(F(x)\), such that \(F'(x) = f(x)\), where \(f(x)\) is our integrand. The result includes a constant of integration \(C\), because when differentiating, the constant vanishes. Thus, there can be infinitely many antiderivatives, each differing by a constant.
In our exercise, the indefinite integration's primary focus is transforming \(\frac{\cos x - 1}{x}\) into its antiderivative form, using a series representation to ease the integration process. Each term in this representation is integrated separately, which simplifies what could be an extremely complex calculation into manageable parts:
When integrating, we seek a function \(F(x)\), such that \(F'(x) = f(x)\), where \(f(x)\) is our integrand. The result includes a constant of integration \(C\), because when differentiating, the constant vanishes. Thus, there can be infinitely many antiderivatives, each differing by a constant.
In our exercise, the indefinite integration's primary focus is transforming \(\frac{\cos x - 1}{x}\) into its antiderivative form, using a series representation to ease the integration process. Each term in this representation is integrated separately, which simplifies what could be an extremely complex calculation into manageable parts:
Term-by-Term Integration
Term-by-term integration allows us to integrate each term in a function's series separately. This approach is particularly useful for series representations like Maclaurin or Taylor series.
Once we have expressed an infinite series, such as the one derived from the Maclaurin series, it is often easiest to integrate by handling each term individually. We perform straightforward integration on each polynomial term.
This method breaks down what might be overwhelming mathematics into more digestible computations that involve basic power rule applications for integration, leading to the series solution for the indefinite integral.
Once we have expressed an infinite series, such as the one derived from the Maclaurin series, it is often easiest to integrate by handling each term individually. We perform straightforward integration on each polynomial term.
- For example, the integral of \( -\frac{x}{2}\) becomes \( -\frac{x^2}{4} + C_1\)
- The integral of \( \frac{x^3}{24}\) turns into \( \frac{x^4}{96} + C_2\)
This method breaks down what might be overwhelming mathematics into more digestible computations that involve basic power rule applications for integration, leading to the series solution for the indefinite integral.
Other exercises in this chapter
Problem 45
Find the value of $$c\( if \)\sum_{n=2}^{\infty}(1+c)^{-n}=2$$
View solution Problem 45
(a) Show that \(\sum_{n=0}^{\infty} x^{n} / n !\) converges for all \(x\). (b) Deduce that \(\lim _{n \rightarrow \infty} x^{n} / n !=0\) for all \(x\).
View solution Problem 46
Find all positive values of \(b\) for which the series \(\sum_{n=1}^{\infty} b^{\ln n}\) converges.
View solution Problem 46
Evaluate the indefinite integral as an infinite series. $$\int \arctan \left(x^{2}\right) d x$$
View solution