/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 Evaluate the following integrals... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
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 by step solution

01

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: $$\int_{0}^{\pi} 2^{\sin x} \cos x d x = 2\int_{0}^{\pi/2} 2^{\sin x} \cos x d x$$
02

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: $$2\int_{0}^{\pi/2} 2^{\sin x} \cos x d x = 2\int_{0}^{1} 2^t dt$$
03

Integrate the function

Now we can easily integrate the function: $$2\int_{0}^{1} 2^t dt = 2 \cdot \left[\frac{2^t}{\ln 2}\right]_{0}^{1}$$
04

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: $$\int_{0}^{\pi} 2^{\sin x} \cos x d x = \frac{2}{\ln 2}$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Logarithm properties Use the integral definition of the natural logarithm to prove that \(\ln (x / y)=\ln x-\ln y\)

Suppose a force of \(15 \mathrm{N}\) is required to stretch and hold a spring \(0.25 \mathrm{m}\) from its equilibrium position. a. Assuming the spring obeys Hooke's law, find the spring constant \(k\) b. How much work is required to compress the spring \(0.2 \mathrm{m}\) from its equilibrium position? c. How much additional work is required to stretch the spring \(0.3 \mathrm{m}\) if it has already been stretched \(0.25 \mathrm{m}\) from its equilibrium position?

Suppose a cylindrical glass with a diameter of \(\frac{1}{12} \mathrm{m}\) and a height of \(\frac{1}{10} \mathrm{m}\) is filled to the brim with a 400-Cal milkshake. If you have a straw that is 1.1 m long (so the top of the straw is \(1 \mathrm{m}\) above the top of the glass), do you burn off all the calories in the milkshake in drinking it? Assume that the density of the milkshake is \(1 \mathrm{g} / \mathrm{cm}^{3}(1 \mathrm{Cal}=4184 \mathrm{J})\)

Two bars of length \(L\) have densities \(\rho_{1}(x)=4 e^{-x}\) and \(\rho_{2}(x)=6 e^{-2 x},\) for \(0 \leq x \leq L\) a. For what values of \(L\) is bar 1 heavier than bar \(2 ?\) b. As the lengths of the bars increase, do their masses increase without bound? Explain.

a. The definition of the inverse hyperbolic cosine is \(y=\cosh ^{-1} x \Leftrightarrow x=\cosh y,\) for \(x \geq 1,0 \leq y<\infty .\) Use implicit differentiation to show that \(\frac{d}{d x}\left(\cosh ^{-1} x\right)=\) \(1 / \sqrt{x^{2}-1}\). b. Differentiate \(\sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1})\) to show that \(\frac{d}{d x}\left(\sinh ^{-1} x\right)=1 / \sqrt{x^{2}+1}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.