Problem 11 - Calculus: An Intuitive and Physical Approach

Question

Evaluate: (a) \(\int \frac{3 x}{x^{2}+2} d x\). Ans. \(\frac{3}{2} \log \left(x^{2}+2\right)+C\). (b) \(\int \frac{x^{2}}{1-x^{3}} d x\). (c) \(\int \frac{x+1}{x^{2}+2 x+5} d x\). (d) \(\int\left(\frac{1}{2 x-5}-\frac{1}{2 x+3}\right) d x\). (e) \(\int \tan \frac{x}{2} d x\).

Step-by-Step Solution

Answer

The solutions are (a) \(\frac{3}{2}\log|x^{2}+2|+C\), (b) \(-\frac{1}{3}\log|1-x^{3}|+C\), (c) \(\frac{1}{2}\arctan\frac{x+1}{2} + C\), (d) \(\frac{1}{2}\log|2x-5| - \frac{1}{2}\log|2x+3| + C\), (e) \(-\log|\cos^2\frac{x}{2}| + C\).

Step 1: Simplify the Integral of 3x/(x^2+2)

This integral is in a form where a simple substitution will work. Let u = x^2 + 2. Then, du = 2x dx. The integral becomes: \(\frac{1}{2}\int\frac{1}{u}du\).

Step 2: Integrate and Substitute Back

The integral of \(\frac{1}{u}\) is \(\ln|u|\), so we get \(\frac{1}{2}\ln|u| + C\). Substituting back, we have \(\frac{3}{2}\ln|x^2+2|+C\).

Step 3: Use Partial Fractions for \(\int \frac{x^2}{1-x^3} dx\)

First notice that the denominator can be written as \((1 - x^3)\). This integral can be approached with partial fractions, and the first step is to express \(\frac{x^2}{1-x^3}\) as a sum of simpler fractions.

Step 4: Find the Anti-Derivative with Substitution

The expression \(\frac{x^2}{1-x^3}\) suggests a substitution. Letting \(u = 1 - x^3\) gives us \(du = -3x^2 dx\), so the integral then becomes \(-\frac{1}{3}\int\frac{1}{u}du\), which integrates to \(-\frac{1}{3}\ln|u| + C\), or ultimately \(-\frac{1}{3}\ln|1-x^3| + C\) after reverting back to x.

Step 5: Integrate \(\int \frac{x+1}{x^2+2x+5} dx\) by Completing the Square

Complete the square in the denominator: \(x^2+2x+5 = (x+1)^2 + 4\). This suggests the substitution \(u = (x+1)\), giving us \(du = dx\). Hence, the integral becomes \(\int\frac{1}{u^2+4}du\).

Step 6: Use a Trigonometric Substitution

Using the identity \(\arctan z = \frac{1}{a} \int\frac{1}{z^2+a^2}dz\) where a=2, we get the integral \(\frac{1}{2}\arctan\frac{u}{2} + C\). Substituting back, we have \(\frac{1}{2}\arctan\frac{x+1}{2} + C\).

Step 7: Decompose the Integral of a Difference of Fractions

This is direct integration after splitting the integral into two. Integrating each term separately by taking \(\int\frac{1}{2x-5}dx\) and \(\int\frac{-1}{2x+3}dx\) respectively.

Step 8: Solve Each Integral by Logarithmic Integration

For the first integral, use substitution: let \(u = 2x-5\). For the second integral, let \(v = 2x+3\). Integrating each gives \(\frac{1}{2}\ln|2x-5|\) and \(-\frac{1}{2}\ln|2x+3|\), respectively. Combining these results in \(\frac{1}{2}\ln|2x-5| - \frac{1}{2}\ln|2x+3| + C\).

Step 9: Integrate \(\int \tan\frac{x}{2} dx\)

Use the trigonometric identity \(\tan z = \frac{\sin z}{\cos z}\), and substitute \(u = \frac{x}{2}\), then \(du = \frac{1}{2}dx\). The integral translates into \(2\int \tan u \, du\).

Step 10: Solve the Integral Using a Trigonometric Identity

Knowing the antiderivative of \(\tan u\) is \(-\ln|\cos u|\), we integrate to get \(-2\ln|\cos\frac{x}{2}| + C\), which can be written, using properties of log, as \(-\ln|\cos^2\frac{x}{2}| + C\).

Key Concepts

Indefinite Integrals

Indefinite integrals represent the collection of all antiderivatives of a function. They are a crucial concept in calculus for finding the original function given its rate of change. For example, if we want to find a function whose derivative is \( 2x \), then we look for its indefinite integral, denoted by \( \int 2x \, dx \), which is \( x^2 + C \), where \(C\) represents the constant of integration. This constant arises because the derivative of a constant is zero, so any constant could have been part of the original function.

Remember that indefinite integrals come with a \( +C \) because of the infinite number of antiderivatives that exist for any given function, due to this additive constant. Approaching an indefinite integral often requires recognizing patterns or using techniques to simplify the expression, as shown in the provided step-by-step solutions for various functions.

Integration by Substitution

Integration by substitution is akin to the chain rule in reverse. It's a valuable technique for evaluating integrals that don't appear to be easily integrable in their current form. Essentially, you choose a part of the integral to substitute with a variable (usually \( u \)), which simplifies the integral into a more familiar form.

Considering example (a), we let \( u = x^2+2 \) and \( du = 2x dx \), which simplifies our integral to \( \frac{1}{2}\int\frac{1}{u} du \), an easier form to integrate directly. The solution then involves finding the integral, in this case, \( \ln|u| \), and substituting back with the original variable to obtain the solution. Substitution is powerful, especially when dealing with composite functions and requires keen observation to identify the right part of the integral to replace with a new variable.

Partial Fraction Decomposition

Partial fraction decomposition is a process used to break down complex rational expressions into simpler ones that are more straightforward to integrate. The goal is to express the integrand as a sum of simpler fractions whose denominators are polynomials of smaller degrees, often linear or quadratic factors.

The provided step-by-step solution involves breaking down the fraction in example (b), which entails expressing the complex fraction as a sum of simpler fractions. This technique is particularly useful when the denominator can be factored into distinct linear or irreducible quadratic factors. Once the fraction is decomposed, integrating each term becomes a much more manageable task.

Trigonometric Integration

Trigonometric integration involves integrating functions with trigonometric identities. It is often used when the integrand involves trigonometric functions such as sine, cosine, or tangent. In such cases, it's helpful to remember key trigonometric identities and use them to simplify the integrand before attempting to integrate.

In example (e), we use the identity for tangent in terms of sine and cosine (\( \tan{u} = \frac{\sin{u}}{\cos{u}} \) ) to transform the integral into a form that allows us to apply a substitution. Understanding and utilizing these trigonometric identities can convert a challenging integral into a standard form, enabling us to find the antiderivative.

Logarithmic Integration

Logarithmic integration involves finding the integral of expressions that can be simplified to the form \( \frac{1}{x} \), whose antiderivative is the natural logarithm of the absolute value of x ( \( \ln|x| \) ). It is commonly employed when dealing with functions that, upon using integration techniques like substitution, yield an integrand proportional to \( \frac{1}{u} \) where \( u \) is some function of \( x \).

As with the function in step 8, the difference of fractions could be integrated by using a simple substitution for each term, and then integrating to get logarithmic expressions for each part. It's essential to handle the absolute value correctly, something that follows from the properties of the logarithm. Summing up these logarithmic functions with their appropriate signs leads to the final answer.

Completing the Square in Integrals

Completing the square is a method used to rewrite quadratic expressions in a form that makes integrals easier to solve. This technique is especially helpful when dealing with quadratics in the denominator of a fraction or within certain trigonometric integrations.

In example (c), \( x^2 + 2x + 5 \) is rewritten as \( (x+1)^2 + 4 \) by completing the square. This new form suggests a substitution that simplifies the integral into one that resembles the inverse tangent function \( \arctan \)—a form that is straightforward to integrate. Completing the square can turn an intimidating integral into one with a known antiderivative, which simplifies finding the solution significantly.