Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-54, perform the indicated integrations. \(\int_{0}^{\pi / 6} 2^{\cos x} \sin x d x\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{2 - 2^{\frac{\sqrt{3}}{2}}}{\ln 2} \).

Step 1: Identify the Integration Technique

The integrand includes the expression \(2^{\cos x} \sin x\). This suggests that a substitution method might be useful. We notice that if we set \(u = \cos x\), then \(du = -\sin x\,dx\). This substitution can simplify the integration.

Step 2: Apply the Substitution

Substitute \(u = \cos x\), giving \(du = -\sin x\,dx \rightarrow -du = \sin x\,dx\). Also, change the limits of integration: when \(x = 0\), \(u = \cos 0 = 1\); when \(x = \frac{\pi}{6}\), \(u = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\). This gives us the integral: \(-\int_{1}^{\frac{\sqrt{3}}{2}} 2^u \, du\).

Step 3: Integrate with Respect to \(u\)

We now integrate \(-\int_{1}^{\frac{\sqrt{3}}{2}} 2^u \, du\). The antiderivative of \(a^u\) is \(\frac{a^u}{\ln a}\). Thus: \[\int 2^u \, du = \frac{2^u}{\ln 2}.\] Substitute back to our integral: \[ -\int_{1}^{\frac{\sqrt{3}}{2}} 2^u \, du = -\left[ \frac{2^u}{\ln 2} \right]_{1}^{\frac{\sqrt{3}}{2}}. \]

Step 4: Evaluate the Definite Integral

Evaluate the expression: \[-\left[\frac{2^u}{\ln 2}\right]_{1}^{\frac{\sqrt{3}}{2}} = -\left( \frac{2^{\frac{\sqrt{3}}{2}}}{\ln 2} - \frac{2^1}{\ln 2} \right) = -\left( \frac{2^{\frac{\sqrt{3}}{2}} - 2}{\ln 2} \right). \] Simplify to get the final result: \[ \frac{2 - 2^{\frac{\sqrt{3}}{2}}}{\ln 2}. \]

Step 5: Conclusion: Write the Final Result

The result of the integration \( \int_{0}^{\pi / 6} 2^{\cos x} \sin x \, dx \) is \( \frac{2 - 2^{\frac{\sqrt{3}}{2}}}{\ln 2} \).

Key Concepts

Substitution Method

The substitution method is a powerful tool when tackling integrals that are challenging or cumbersome. It helps by transforming a difficult integral into a simpler one.

Here's the basic idea:

• You identify a part of the integrand (the function being integrated) that can be replaced with a single variable, usually denoted as 'u'. This process is called substitution.
• This substitution simplifies the integral into a more manageable form.
• Along with this variable change, you must also transform the differential (in this exercise, from 'dx' to 'du').

In this specific problem, the substitution chosen was:

• For the function, choose: = cos x
• Thus, the differential becomes: du = -sin x dx

This substitution replaced the original integral ewline ewlineThe limits of integration were also changed based on the function values at the bounds x = 0 and x = \(\frac{\pi}{6}\). This effectively changes the problem from being in terms of x, to being in terms of u, simplifying the process.

Definite Integrals

Definite integrals are a fundamental concept in calculus, providing a way to calculate the net area under a curve. These integrals are evaluated over a specific interval, which is defined by upper and lower bounds.

They are written in the form: ewlineewline\(\int_{a}^{b} f(x) \, dx\)ul>

The expression within the integral sign, f(x), is the function being integrated.

The limits of integration at the bottom (a) and top (b) of the integral sign indicate the interval over which you are integrating.

The primary goal of a definite integral is to find the accumulated value of the function over the specified range of integration. In the present exercise, after performing the substitution and integrating with respect to u, we evaluated the definite integral by substituting back the transformed upper and lower limits of \(\int_{1}^{\frac{\sqrt{3}}{2}} 2^u \, du \) back into the original function. Evaluating these at the given bounds yielded the result. This detailed understanding helps ensure that every step is systematically approached with precision and accuracy.

Exponential Functions

Exponential functions are often encountered in calculus problems and understanding them is crucial for solving integrals involving them.

In these functions, the variable is in the exponent of a constant base. They are expressed in the form:e^x or b^xThe key characteristics of these functions include:

• They grow rapidly compared to polynomial functions, especially as the exponent increases.
• They have constant relative rates of change, which means they grow by consistent percentages over equal increments.

In the exercise we're discussing, we specifically dealt with an expression with base 2:

• The expression integrated was \(2^{\cos x}\), emphasizing a variable exponent.
• One crucial aspect here is the differentiation of exponentials, where the derivative of \(a^x\) is \(a^x \ln a\).
• Understanding the integral of exponentials follows a similar logic, helping to resolve \(\int 2^u \, du\) as noted: \(\frac{2^u}{\ln 2}\).

This shows the importance of understanding not just exponential growth, but the underlying principles and calculations. This approach ensures deeper comprehension, enabling one to tackle varied exponential integration problems.