Problem 62 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{0}^{\pi} 2^{\sin x} \cos x d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $\int_{0}^{\pi} 2^{\sin x} \cos x d x$. Answer: The value of the definite integral is $\frac{2}{\ln 2}$.

Step 1: Observe the function and use the property for definite integral

The integral given is symmetric with respect to the line \(x=\frac{\pi}{2}\). So we can rewrite it by using the property of symmetry for definite integrals: $$\int_{0}^{\pi} 2^{\sin x} \cos x d x = 2\int_{0}^{\pi/2} 2^{\sin x} \cos x d x$$

Step 2: Substitute trigonometric function

Now, we'll use substitution method to make it easier to evaluate. Let \(t = \sin x\). Then, \(dt = \cos x d x\). Now substitute \(t\) and \(dt\) into the integral and adjust the boundaries accordingly: $$2\int_{0}^{\pi/2} 2^{\sin x} \cos x d x = 2\int_{0}^{1} 2^t dt$$

Step 3: Integrate the function

Now we can easily integrate the function: $$2\int_{0}^{1} 2^t dt = 2 \cdot \left[\frac{2^t}{\ln 2}\right]_{0}^{1}$$

Step 4: Evaluate the definite integral

To evaluate the definite integral, substitute the limits of integration and subtract: $$2 \cdot \left[\frac{2^t}{\ln 2}\right]_{0}^{1} = 2 \cdot \left[\frac{2^{1}}{\ln 2} - \frac{2^{0}}{\ln 2}\right] = 2 \cdot \left[\frac{2}{\ln 2} -\frac{1}{\ln 2}\right] = \frac{2}{\ln 2}$$ So, the result of the integral is: $$\int_{0}^{\pi} 2^{\sin x} \cos x d x = \frac{2}{\ln 2}$$

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a clever technique used in integral calculus to simplify the integration process. By substituting trigonometric functions for variables, we can often transform complex integrals into more manageable forms. In the exercise, trigonometric substitution is effectively used by setting \( t = \sin x \). This substitution changes the variable of integration from \( x \) to \( t \), making the integral simpler to solve.

When we make this substitution, we also need to change the differential element correspondingly. Here, \( dt = \cos x \, dx \) is derived from \( t = \sin x \). It simplifies the integral because after substitution, the function becomes \( 2^t \), a much easier expression to integrate over the new limits of \( t \).

Switching to trigonometric terms often aligns the integral with common trigonometric identities and integrals, allowing for easier integration and further simplification.

Symmetry in Integrals

Symmetry properties can significantly simplify definite integrals, especially when the function is symmetric with respect to the limits. A function having "symmetric properties" means that its behavior across a certain point - in this case, the line \( x=\frac{\pi}{2} \) - is mirrored.

The problem takes advantage of this symmetry to simplify the definite integral. The original integral from \( 0 \) to \( \pi \) can be split and rewritten using the symmetry property: \( \int_{0}^{\pi} 2^{\sin x} \cos x \, dx = 2 \int_{0}^{\pi/2} 2^{\sin x} \cos x \, dx \). This reduces the integral to half its original limits, simplifying computation while ensuring correctness of the result.

Using symmetry helps in reducing computational complexity and reveals properties of the function that can be exploited to obtain results more efficiently.

Integration by Substitution

Integration by substitution is a powerful technique for solving integrals, very much akin to the reverse of the chain rule for derivatives. This method is particularly useful when dealing with composite functions, allowing for transformation into simpler, solvable integrals.

In the given exercise, after using symmetry and trigonometric substitution, the integral \( \int_{0}^{1} 2^t \, dt \) is set up for easy evaluation through integration by substitution. Here, substituting \( t \) simplifies the integral to an exponential form, \( 2^t \), which is straightforward to integrate.

The result of integrating \( 2^t \) is expressed as \( \frac{2^t}{\ln 2} \). The bounds \( [0, 1] \) are then applied to compute the definite integral, \( \left[ \frac{2}{\ln 2} - \frac{1}{\ln 2} \right] \), yielding the final result \( \frac{2}{\ln 2} \). Integration by substitution not only simplifies the expression but also clarifies the pathway to solving the integral effectively.