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Evaluate the integrals using appropriate substitutions. $$ \int \frac{d x}{1+16 x^{2}} $$

Short Answer

Expert verified
\( \int \frac{dx}{1 + 16x^2} = \frac{1}{4} \arctan(4x) + C \)

Step by step solution

01

Identify the Form of the Integral

The integral is of the form \( \int \frac{dx}{1 + (ax)^2} \), which suggests a trigonometric substitution. Specifically, this resembles the integral involving \( \arctan \).
02

Choose the Appropriate Substitution

To simplify the integral, use the substitution \( x = \frac{1}{4} \tan(\theta) \). This substitution transforms \( 16x^2 \) into \( \tan^2(\theta) \).
03

Differentiate the Substitution

Differentiate the substitution to find \( dx \) in terms of \( d\theta \). With \( x = \frac{1}{4} \tan(\theta) \), we have:\[ dx = \frac{1}{4} \sec^2(\theta) \, d\theta \]
04

Rewrite the Integral in Terms of \( \theta \)

Substitute \( x = \frac{1}{4} \tan(\theta) \) and \( dx = \frac{1}{4} \sec^2(\theta) \, d\theta \) into the integral:\[ \int \frac{dx}{1 + 16x^2} = \int \frac{\frac{1}{4} \sec^2(\theta) \, d\theta}{1 + \tan^2(\theta)} \]
05

Simplify Using Trigonometric Identity

Recall the trigonometric identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \). Substitute this into the integral:\[ \int \frac{\frac{1}{4} \sec^2(\theta) \, d\theta}{\sec^2(\theta)} = \int \frac{1}{4} \, d\theta \]
06

Integrate and Back-Substitute

The integral becomes:\[ \frac{1}{4} \int d\theta = \frac{1}{4} \theta + C \]Since \( \theta = \arctan(4x) \), back-substitute to get:\[ \frac{1}{4} \arctan(4x) + C \]
07

Define the Final Answer

Combine all simplifications and substitutions to define the final result for the integral:\[ \int \frac{dx}{1 + 16x^2} = \frac{1}{4} \arctan(4x) + C \]

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Key Concepts

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

Integral Calculus
Integral calculus is a major branch of mathematics that deals with finding the antiderivative or integral of functions. It is essentially about measuring the area under a curve described by a function. This area translates to many practical applications including physics, engineering, and statistics. Integrals can be definite, which means they have limits and provide a numerical answer, or indefinite, like in the exercise, where they represent a family of functions.

When solving integrals, we are often seeking a function whose derivative yields the original function under the integral sign. This requires various methods and techniques, depending on the form of the function we are integrating. The exercise demonstrates an indefinite integral, showing us how we can approach solving for such integrals using different techniques and substitutions. For instance, the technique used in the exercise, trigonometric substitution, is one of those essential strategies.
Integration Techniques
Integration techniques are essential tools that allow us to solve integrals which may not be immediately solvable by basic integration rules. These techniques transform complex integrals into simpler forms. One of the techniques used in the exercise is **trigonometric substitution**.

Trigonometric substitution is particularly useful when the integral contains expressions like \(x^2 + a^2\), \(x^2 - a^2\), or \(a^2 - x^2\). In these cases, trigonometric identities can simplify such expressions since they naturally involve squared terms. For instance, substituting \(x = \frac{1}{4} \tan(\theta)\) transformed the integral into a form that uses the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\), greatly simplifying the integration process.

This specific substitution also changes the variable, requiring us to find \(dx\) in terms of \(d\theta\) using differentiation, which is a vital step to correctly apply this technique.
Antiderivatives
Antiderivatives, also known as indefinite integrals, are functions that represent the 'reverse' of differentiation. If you have a function \(F(x)\), and its derivative \(F'(x)\) equals the function \(f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\).

In the context of the exercise, finding the antiderivative involves deciphering the function inside the integral and transforming it into a simpler problem that can be easily integrated. Here, the process transforms the complicated fraction into a simpler trigonometric problem, \((1/4)\theta\), where \(\theta = \arctan(4x)\). This makes it easy to integrate to \(\frac{1}{4} \theta + C\).

The constant \(C\) is included because the integration process is indefinite. It accounts for any constants that would disappear during differentiation. Conclusively, the antiderivative represents a whole family of functions that between them make up all possible solutions to the integral problem.

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Most popular questions from this chapter

$$ \int \frac{y}{\sqrt{2 y+1}} d y $$

$$ \int \sec ^{3} 2 x \tan 2 x d x $$

$$ \int e^{\sin x} \cos x d x $$

Electricity is supplied to homes in the form of alternating current, which means that the voltage has a sinusoidal waveform described by an equation of the form $$ V=V_{p} \sin (2 \pi f t) $$ (see the accompanying figure). In this equation, \(V_{p}\) is called the peak voltage or amplitude of the current, \(f\) is called its frequency, and \(1 / f\) is called its period. The voltages \(V\) and \(V_{p}\) are measured in volts \((\mathrm{V})\), the time \(t\) is measured in seconds (s), and the frequency is measured in hertz (Hz). \((1 \mathrm{~Hz}=1\) cycle per second; a cycle is the electrical term for one period of the waveform.) Most alternating-current voltmeters read what is called the \(\boldsymbol{r m s}\) or root-mean-square value of \(V\). By definition, this is the square root of the average value of \(V^{2}\) over one period. (a) Show that $$ V_{\mathrm{rms}}=\frac{V_{p}}{\sqrt{2}} $$ [Hint: Compute the average over the cycle from \(t=0\) to \(t=1 / f\), and use the identity \(\sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta)\) to help evaluate the integral.] (b) In the United States, electrical outlets supply alternating current with an rms voltage of \(120 \mathrm{~V}\) at a frequency of \(60 \mathrm{~Hz}\). What is the peak voltage at such an outlet?

Evaluate each integral by first modifying the form of the integrand and then making an appropriate substitution, if needed $$ \int\left[\ln \left(e^{x}\right)+\ln \left(e^{-x}\right)\right] d x $$

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