Problem 53 - Calculus

Question

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

Step-by-Step Solution

Answer

The integral evaluates to 0 because the limits remain the same after substitution.

Step 1: Identify Integration Technique

We need to evaluate the integral \( \int_{0}^{\pi} 2^{\sin x} \cos x \, dx \). Notice that \( 2^{\sin x} \) can be challenging to integrate directly, but the presence of \( \cos x \) suggests a possible substitution. Let's use the substitution \( u = \sin x \).

Step 2: Substitute Variables

Let \( u = \sin x \), which gives \( du = \cos x \, dx \). This allows us to rewrite the integral in terms of \( u \) as \( \int 2^u \, du \).

Step 3: Change Limits of Integration

When \( x = 0 \), \( u = \sin 0 = 0 \). When \( x = \pi \), \( u = \sin \pi = 0 \). Hence, the limits of integration become from 0 to 0.

Step 4: Evaluate the Integral

Since the definite integral \( \int_{0}^{0} 2^u \, du \) has identical upper and lower limits, its value is automatically zero.

Step 5: Conclude the Process

Therefore, the evaluated integral \( \int_{0}^{\pi} 2^{\sin x} \cos x \, dx \) equals zero.

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus for simplifying the process of integration. When faced with a complex integral, like the one involving trigonometric functions and arithmetic operations in the exercise, substitution can make the problem more manageable. Here's what you typically do:

• You start by identifying a part of the integral to substitute with a new variable, often denoted as \(u\).
• This usually involves choosing \(u\) such that the derivative \(du\) also appears in the integral, simplifying the entire expression.
• In the original exercise, this was evident when we substituted \(u = \sin x\), thanks to the presence of \(\cos x\, dx\). This makes \(du = \cos x\, dx\).

This approach not only transforms the integral into a simpler form but often helps in switching the integral into a function that is easier to evaluate directly. It's akin to translating a problem into a simpler language you can understand.

Limits of Integration

Switching the variable of integration affects the limits of integration as well. This step is crucial since it dictates the values you compute your final result over. In the substitution method:

• As you substitute \(x\) with \(u\), you must also change the limits to reflect this new variable.
• In our example, initially, the limits for \(x\) were from 0 to \(\pi\).
• Upon substitution, we calculated the corresponding \(u\) limits. When \(x=0\), \(\sin 0 = 0\), and similarly, when \(x = \pi\), \(\sin \pi = 0\). Thus, the new limits for \(u\) became 0 to 0.

This change inherently simplified the integration since both limits were the same, transforming the integral into one for which the result is instantly zero. This is a classic scenario in definite integrals where the same limits imply no area under the curve is calculated.

Trigonometric Functions

Trigonometric functions appear frequently in calculus, often leading to integrals that need clever techniques to solve. In the original integral, we dealt with \(2^{\sin x}\) and \(\cos x\), both trigonometrically significant:

• The function \(\sin x\) was initially cumbersome due to \(2^{\sin x}\), a composition outside typical formulas.
• The \(\cos x\) conveniently provided the derivative needed for the substitution method, transforming this problem.
• Recognizing these functions and how they relate to each other is key, as often one function's derivative is another in the same family of trigonometric identities.

Trigonometric identities and functions are fundamental in transforming and solving integrals. Understanding their relationships can often simplify evaluation and is why substitution is particularly effective with these forms of integrals.