Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

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: Recognize the Integration by Substitution

First, notice that the integral \( \int 2^{\cos x} \sin x \, dx \) might be suitable for substitution. The derivative of \( \cos x \) is \(-\sin x\), which corresponds to the \( \sin x \) term in the integral.

Step 2: Choose a Substitution

Let's set \( u = \cos x \). Then, the derivative \( du = -\sin x \, dx \). This implies that \(-du = \sin x \, dx \). Substitute \( u \) and \( du \) into the integral.

Step 3: Substitute and Change Limits

Substitute \( u \) and \( du \) into the integral, replacing \( \sin x \, dx \) with \(-du \), so the integral becomes \[ -\int 2^u \, du \]. Change the limits of integration as well: - When \( x = 0 \), \( u = \cos(0) = 1 \) - When \( x = \frac{\pi}{6} \), \( u = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \). Thus, the limits change from \( 0 \) to \( 1 \) and from \( \frac{\pi}{6} \) to \( \frac{\sqrt{3}}{2} \).

Step 4: Integrate with Respect to \( u \)

You now have the integral:\[ -\int_{1}^{\frac{\sqrt{3}}{2}} 2^u \, du \].The integral of \( 2^u \) with respect to \( u \) is \( \frac{2^u}{\ln(2)} \). Applying this gives:\[ -\left[ \frac{2^u}{\ln(2)} \right]_{1}^{\frac{\sqrt{3}}{2}} \].

Step 5: Apply the Fundamental Theorem of Calculus

Evaluate the definite integral using the antiderivative:\[ = -\left( \frac{2^{\frac{\sqrt{3}}{2}}}{\ln(2)} - \frac{2^1}{\ln(2)} \right) \].Simplify this to:\[ = -\left( \frac{2^{\frac{\sqrt{3}}{2}} - 2}{\ln(2)} \right) \].Simplifying further, the result is:\[ = \frac{2 - 2^{\frac{\sqrt{3}}{2}}}{\ln(2)} \].

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique often used in calculus to simplify integrals. The idea behind this method is to transform a complicated integral into a more manageable form.
By performing a clever substitution of variables, you can make the integration process much easier.
In this exercise, the choice of substitution was key to solving the problem:

• We set \( u = \cos x \), which simplified the expression significantly.
• The derivative \( du = -\sin x \, dx \) allowed us to express \( \sin x \, dx \) as \(-du\).
• This substitution nicely transformed the integral into a basic exponential form \( -\int 2^u \, du \).

Integration by substitution is an essential skill for tackling a variety of complicated integrals.
By choosing the right substitution, you can ease the integration process and solve the problem efficiently.

Definite Integral

A definite integral represents the accumulation of quantities and can be viewed as the net area under a curve.
In this problem, we started with \( \int_{0}^{\pi / 6} 2^{\cos x} \sin x \, dx \), which involves integrating between the limits \( 0 \) and \( \pi/6 \).
After substitution, the limits of the integral changed as well:

• When \( x = 0 \), \( u = \cos(0) = 1 \).
• When \( x = \pi/6 \), \( u = \cos(\pi/6) = \frac{\sqrt{3}}{2} \).

These boundaries ensure that the integral is evaluated over the correct range.
Definite integrals are crucial in finding areas, solving physics problems, and assessing changes in quantities from one state to another.

Trigonometric Functions

Trigonometric functions are fundamental in calculus, especially when dealing with periodic processes and waves.
In this integration problem, the trigonometric function involved was \( \cos x \).
By recognizing and using the properties of trigonometric functions:

• We exploited the function \( \sin x \) to guide our choice of substitution.
• We understood the cosine function's role in determining substitution limits and derivatives.

Trigonometric integration often requires familiarity with identities and derivatives.
Understanding these functions helps in executing substitutions effectively and is critical for more advanced calculus applications.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration.
It includes two main parts:

• The first part relates the derivative to the integral, establishing that differentiation and integration are reverse processes.
• The second part provides the method for evaluating definite integrals. It states that if \( F \) is an antiderivative of \( f \) on an interval, then \( \int_a^b f(x) \, dx = F(b) - F(a) \).

In this exercise, after integrating with respect to \( u \), we used this theorem to evaluate the definite integral:

• The antiderivative \( \frac{2^u}{\ln(2)} \) was first calculated.
• The definite integral was then evaluated by substituting the upper and lower limits which resulted in the expression \( \frac{2 - 2^{\frac{\sqrt{3}}{2}}}{\ln(2)} \).

This theorem is a cornerstone of calculus, allowing for the precise computation of definite integrals and relating them to antiderivatives.