Problem 68
Question
Evaluate the integrals in Exercises \(47-70\) $$ \int_{1}^{e^{\pi / 4}} \frac{4 d t}{t\left(1+\ln ^{2} t\right)} $$
Step-by-Step Solution
Verified Answer
The evaluated integral is \( \pi \).
1Step 1: Identify the Integral
The given integral is \( \int_{1}^{e^{\pi / 4}} \frac{4 \, dt}{t(1+\ln^{2} t)} \). We need to evaluate this definite integral from \( t = 1 \) to \( t = e^{\pi / 4} \).
2Step 2: Choose a Substitution
To simplify the integration, use the substitution \( u = \ln t \). Then \( du = \frac{1}{t} dt \). The limits of integration change accordingly: when \( t = 1 \), \( u = 0 \), and when \( t = e^{\pi / 4} \), \( u = \frac{\pi}{4} \).
3Step 3: Rewrite the Integral
Substitute \( u \) and \( du \) into the integral. The integral becomes:\[\int_{0}^{\pi/4} \frac{4 \, du}{1+u^2}\]
4Step 4: Recognize the Integral Form
The integral \( \int \frac{du}{1+u^2} \) is known to be \( \arctan u + C \). Thus, we recognize that this integral is a standard form that results in the arctangent function.
5Step 5: Evaluate the Integral
Evaluate the integral:\[\int_{0}^{\pi/4} \frac{4 \, du}{1+u^2} = 4 \left[ \arctan(u) \right]_{0}^{\pi/4}\]This becomes:\[4 \left( \arctan(\pi/4) - \arctan(0) \right)\]
6Step 6: Calculate the Result
Calculate the arctangent values:- \( \arctan(\pi/4) \approx \frac{\pi}{4} \) (since \( \pi/4 \) is the angle whose tangent is \( \pi/4 \)).- \( \arctan(0) = 0 \).So, the result is:\[4 \left( \frac{\pi}{4} - 0 \right) = \pi\]
Key Concepts
Substitution MethodIntegration LimitsArctangent FunctionChange of Variable
Substitution Method
The substitution method is a powerful technique used in calculus to simplify the process of integration. By changing the variable of integration, it often transforms a complex integral into a more manageable form. In this exercise, we used the substitution \( u = \ln t \). This particular substitution is helpful because it simplifies the expression involving the natural logarithm. When employing substitution, always remember to:
- Choose a new variable, generally represented as \( u \), that simplifies the integral.
- Express \( dt \) in terms of \( du \) using the derivative of the substitution equation, in this case, \( du = \frac{1}{t} dt \).
- Adjust your integration limits according to the new variable.
Integration Limits
When performing definite integration, it is essential to adjust the integration limits to match the variable you are currently using. The original limits \( t = 1 \) to \( t = e^{\pi / 4} \) were suitable for the variable \( t \). However, after substitution, we switched to the variable \( u \).To update the limits:
- Substitute each original limit into the substitution formula. For example, when \( t = 1 \), substituting gives \( u = \ln(1) = 0 \).
- Do the same with the upper limit: when \( t = e^{\pi / 4} \), substituting gives \( u = \ln(e^{\pi / 4}) = \frac{\pi}{4} \).
Arctangent Function
The arctangent, denoted as \( \arctan(x) \), is a fundamental function in calculus, specifically in integration. It is the inverse function of the tangent and is used when evaluating integrals of the form \( \int \frac{1}{1+u^2} \, du \). This is a standard integral that results in the function \( \arctan(u) \).Here’s how it works:
- The integral \( \int \frac{1}{1+u^2} \, du \) directly transforms into \( \arctan(u) + C \), where \( C \) is the constant of integration.
- This transformation is due to the derivative of the arctangent function: \( \frac{d}{du} (\arctan(u)) = \frac{1}{1+u^2} \).
Change of Variable
A change of variable is a technique used to rewrite an integral in terms of a different variable, often simplifying the integral. In this exercise, changing from \( t \) to \( u \) made the integration much simpler.Considerations for a successful change of variable:
- Choose a substitution that simplifies the integrand to a more recognizable form.
- Express all parts of the integral in terms of the new variable, including converting the differential and adjusting integration limits.
- Look for standard integral forms that are easier to compute, such as the arctangent function in this example.
Other exercises in this chapter
Problem 67
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