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Evaluate the following integrals. Include absolute values only when needed. $$\int_{0}^{\pi} 2^{\sin x} \cos x d x$$

Short Answer

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

Step by step solution

01

Identify the substitution

In this integral, we can use the following substitution: Let \(u = \sin x\), so that \(d u = \cos x d x\). The integral becomes: $$\int 2^u du$$
02

Replace the boundary limits

Since we are using substitution, we need to change the boundary limits of the integral according to \(u\). At \(x=0\), \(u=\sin(x)=\sin(0)=0\). At \(x=\pi\), \(u=\sin(x)=\sin(\pi)=0\). Therefore, the new boundaries are from \(0\) to \(0\). Let's replace these limits in the integral: $$\int_{0}^{0} 2^u du$$
03

Evaluate the integral

The given definite integral is: $$\int_{0}^{0} 2^u du$$ We notice that the lower and upper boundaries are the same. Due to this property, the result of the integral is 0. Hence, the definite integral is: $$\int_{0}^{\pi} 2^{\sin x} \cos x d x = 0$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Substitution
One of the most powerful techniques in calculus to tackle complex integrals is Integration by Substitution. In essence, this method involves replacing a part of the integral with a new variable, simplifying the integral in the process. Think of it as a strategic trade, where a difficult expression is exchanged for a simpler one that is easier to integrate.

Using our example, we have the integral \[ \int_{0}^{\pi} 2^{\sin x} \cos x dx \]. The substitution made is \( u = \sin x \), transforming the difficult-to-integrate expression into \( 2^u \), with \( du = \cos x dx \). This change of variable simplifies the integration process, converting a trigonometric integral into a straightforward exponential integral.

The skill lies in recognizing patterns and selecting the most appropriate substitution. Properly executed, this technique reduces the integral into a form that not only is simpler but also reveals easier paths for finding the integral boundaries, which leads us smoothly into our next concept.
Trigonometric Integration
The study of Trigonometric Integration incorporates the integration of trigonometric functions and is a significant part of calculus. In many cases, these integrals appear complex due to their oscillatory nature; however, strategies like substitution can simplify them significantly.

In the exercise, we initially face an integral containing the function \( 2^{\sin x} \), which is compounded by the presence of the trigonometric function \( \sin x \), making direct integration not straightforward. Yet, by recognizing that \( \cos x dx \) resembles the derivative of \( \sin x \), we set up for integration by substitution.

Once the substitution is applied to a trigonometric function, it’s critical to adjust your mindset to treat the integral as if it were non-trigonometric. This perspective shift simplifies the cognitive load and allows for a more direct route to the integration result.
Integral Boundaries
Understanding Integral Boundaries is crucial when working with definite integrals. These boundaries, often referred to as the limits of integration, dictate the interval over which the function is integrated and can greatly affect the result.

In our exercise, one might expect a complex evaluation process due to the function \( 2^{\sin x} \); however, due to the definite integral's boundaries from \( 0 \) to \( \pi \), we get a surprising twist. When substituting these boundaries into our new variable \( u \), we find that both result in \( u = 0 \). Thus, we arrive at an integral from \( 0 \) to \( 0 \), which mathematically is equal to zero.

The boundaries transformed the potentially complex integral into a simple calculation. It’s essential to always remember to adjust the boundaries when applying substitution and to recognize that the integral of any function over an interval where the boundaries are the same is always zero. These seemingly small details can lead to significant insights and simplify the process of integration.

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