Problem 7 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

nd the inde nite integrals. (a) \(\int \frac{t}{\left(t^{2}+1\right)^{2}} d t\) (b) \(\int \tan 2 t d t\) (c) \(\int \frac{e^{w}}{e^{w}+1} d w\) (d) \(\int \frac{e^{w}}{e^{2 w}+1} d w\)

Step-by-Step Solution

Answer

The solutions to the four integrals are: (a) \( \ln|u| - \frac{1}{u} + C \), (b) \( -\frac{1}{2} \ln|\cos(2t)| + C \), (c) \( \ln|1 + e^{-w}| + C \), and (d) \( \tan^{-1}(e^w) + C \)

Step 1: Problem A Solution

Integral Problem: \( \int \frac{t}{\left(t^{2}+1\right)^{2}} d t \) \n Select the substitution method for this problem. First, make substitution \( u = t^2 + 1 \). Then, the integral becomes \( \int \frac{u-1}{u^2} du \). This can be broken down further to \( \int (u^{-1} - u^{-2}) du \), and then integrated again.

Step 2: Problem B Solution

Integral Problem: \( \int \tan 2 t d t \) \n Apply a substitution of \( u = 2t \). The integral then becomes \( \frac{1}{2} \int \tan u du \). Apply the knowledge that the integral of \( tan(u) \) is \( -ln|cos(u)| \). Substitute \( u \) back into the function to get the final answer.

Step 3: Problem C Solution

Integral Problem: \( \int \frac{e^{w}}{e^{w}+1} d w \)\n Reformulate the original equation as \( \int \frac{e^{w}}{e^{w}} \frac{1}{\frac{1}{e^{w}}+1} dw \). It simplifies to \( \int \frac{1}{1+e^{-w}} dw \). Now make the substitution \( u = 1 + e^{-w} \). Find \( du \) in terms of \( dw \) and then integrate.

Step 4: Problem D Solution

Integral Problem: \( \int \frac{e^{w}}{e^{2 w}+1} d w \)\n Simplify it to \( \int \frac{1}{e^{w} + e^{-w}} dw \). Use the substitution \( u = e^w \), take the derivative with respect to \( w \) and find \( du \). Substitute back into the integral.

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus designed to simplify integration by changing variables. It's particularly useful when the integral isn't straightforward. The key idea is to transform the integrand into a simpler form.
To apply substitution:

• Identify a part of the integrand that allows the transformation of the integral into something easier to integrate. For example, replacing a complex expression with a simpler variable, like letting \( u = g(x) \).
• Differentiating \( u \) gives \( du = g'(x)dx \). Substitute these into the original integral.
• Perform the integration with respect to \( u \), then reverse the substitution to express the final result in terms of the original variable.

By mastering the substitution method, students can tackle a wide variety of calculus problems efficiently.

Integration Techniques

Integration techniques are methods used to find the antiderivatives of functions. These methods are crucial because not all integrals are simple or direct. Several techniques can simplify challenging integrals:

• Substitution Method: As detailed earlier, this is used to simplify the integrand by introducing a new variable.
• Integration by Parts: Useful for products of functions, using the formula \( \int u \, dv = uv - \int v \, du \).
• Partial Fraction Decomposition: Breaks down rational functions into simpler fractions easier to integrate.

Each technique has its appropriate application scenario, and often, a problem may require a combination of methods to solve.

Calculus Problems

In calculus, problems often involve finding the area under curves, which leads to solving integrals. These problems range from simple to complex, requiring various integration techniques. Here’s a breakdown of common calculus problem scenarios:

• Basic Integration: Direct evaluation of simpler functions like polynomials.
• Trigonometric Integration: Such as \( \int \tan(2t) dt \), where substitution and known antiderivatives of trigonometric functions are used.
• Exponential Functions: Like \( \int \frac{e^{w}}{e^{2w} + 1} dw \), often requiring manipulation and substitution for integration.

Approaching calculus problems effectively involves understanding which technique to apply and being familiar with function behavior.

Step-by-step Integration

Step-by-step integration is a logical and organized approach to solving integrals. It allows students to break down complex integrals into manageable parts, checking each step along the way. Here's how it’s done:

• Identify the integrand: Start by thoroughly understanding the function to be integrated.
• Choose a method: Decide on an integration technique that fits the problem best, considering substitution, parts, or another technique.
• Work in phases: Tackle the problem step by step, often transforming it into simpler integrals.
• Check your work: After integrating, differentiate the result to ensure it matches the original integrand.

By following a step-by-step methodology, students not only arrive at the correct solution but also reinforce their understanding of calculus principles.