Chapter 7: Problem 5
Evaluate the integrals. $$\int \frac{3 \sec ^{2} t}{6+3 \tan t} d t$$
Short Answer
Expert verified
\(\ln|2 + \tan t| + C\)
Step by step solution
01
Simplify the Integral
First, notice that the integral can be simplified by factoring out a common factor in the denominator. The given integral is \( \int \frac{3 \sec^2 t}{6 + 3 \tan t} \, dt \). You can factor a 3 out of the denominator to get: \( \int \frac{3 \sec^2 t}{3(2 + \tan t)} \, dt \), which simplifies further to \( \int \frac{\sec^2 t}{2 + \tan t} \, dt \).
02
Substitution
Use substitution to simplify the integration process. Let \( u = \tan t \). Then, the derivative \( du = \sec^2 t \, dt \). Now, substitute \( u \) and \( du \) in the integral: \[ \int \frac{\sec^2 t}{2 + \tan t} \, dt = \int \frac{1}{2 + u} \, du \].
03
Integrate with Respect to u
The integral \( \int \frac{1}{2 + u} \, du \) is a simple logarithmic integral. The antiderivative is \( \ln|2 + u| + C \), where \( C \) is the constant of integration.
04
Substitute Back for t
Replace \( u \) with \( \tan t \) as per our substitution, which results in the expression: \( \ln|2 + \tan t| + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a fundamental concept in integral calculus. It's used to simplify integrals by changing variables. This technique allows us to transform a complex expression into something more manageable. The core idea is to pick a new variable, often denoted as \( u \), which will replace a part of the integrand. By doing this, we can turn a difficult integral into a simpler one.
In our example, the integral is \( \int \frac{\sec^2 t}{2 + \tan t} \, dt \). We let \( u = \tan t \). As a result, the derivative of \( u \) with respect to \( t \) becomes \( du = \sec^2 t \, dt \). This substitution is strategic because it allows us to rewrite the integral in terms of \( u \), making it easier to solve.
Here’s a quick bullet list of how substitution works:
In our example, the integral is \( \int \frac{\sec^2 t}{2 + \tan t} \, dt \). We let \( u = \tan t \). As a result, the derivative of \( u \) with respect to \( t \) becomes \( du = \sec^2 t \, dt \). This substitution is strategic because it allows us to rewrite the integral in terms of \( u \), making it easier to solve.
Here’s a quick bullet list of how substitution works:
- Identify part of the integrand for substitution.
- Express the differential \( dt \) in terms of \( du \).
- Replace all \( t \) expressions with \( u \) equivalents in the integral.
- Solve the simpler integral.
- Substitute back the original variable.
Logarithmic Integral
A logarithmic integral arises when the antiderivative of a function includes a natural logarithm. It typically appears in integrals of the form \( \int \frac{1}{x} \, dx \), whose antiderivative is \( \ln|x| + C \). This form occurs due to the derivative of the natural logarithm function.
In our integral example, after applying the substitution, we ended up with \( \int \frac{1}{2 + u} \, du \). Recognizing this as a logarithmic integral, we can find its antiderivative as \( \ln|2 + u| + C \). This process involves the understanding that dividing by \( (2+u) \) and integrating with respect to \( u \) naturally leads to the logarithmic form, owing to the fundamental properties of integrals and derivatives.
These properties include:
In our integral example, after applying the substitution, we ended up with \( \int \frac{1}{2 + u} \, du \). Recognizing this as a logarithmic integral, we can find its antiderivative as \( \ln|2 + u| + C \). This process involves the understanding that dividing by \( (2+u) \) and integrating with respect to \( u \) naturally leads to the logarithmic form, owing to the fundamental properties of integrals and derivatives.
These properties include:
- The integral of \( \frac{1}{x} \) results in a logarithm.
- Logarithms represent ratios and growth rates.
- The constant \( C \) is always added, representing the general solution to the integral.
Antiderivative
The antiderivative of a function, also known as the indefinite integral, is a function whose derivative is the given function. Finding an antiderivative is essentially the reverse of differentiation. When we integrate a function, we are looking for another function whose derivative will be the integral's original function.
In our example, the integral \( \int \frac{1}{2 + u} \, du \) solves to \( \ln|2 + u| + C \). This is the antiderivative because if you were to differentiate \( \ln|2 + u| \), you'd get back \( \frac{1}{2 + u} \). The constant \( C \) is included because indefinite integrals represent a family of functions, each differing by a constant.
Here's a brief summary of antiderivatives:
In our example, the integral \( \int \frac{1}{2 + u} \, du \) solves to \( \ln|2 + u| + C \). This is the antiderivative because if you were to differentiate \( \ln|2 + u| \), you'd get back \( \frac{1}{2 + u} \). The constant \( C \) is included because indefinite integrals represent a family of functions, each differing by a constant.
Here's a brief summary of antiderivatives:
- They are solutions to indefinite integrals.
- They include a constant \( C \) due to the nature of integration.
- They reverse differentiation.
- The derivative of an antiderivative yields the original function.