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Evaluate the following integrals. $$\int \frac{d x}{\left(16-x^{2}\right)^{1 / 2}}$$

Short Answer

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Question: Evaluate the following integral: $$\int \frac{dx}{\sqrt{16 - x^2}}$$ Answer: The integral can be evaluated as $$\int \frac{dx}{\sqrt{16 - x^2}} = \arcsin\left(\frac{x}{4}\right) + C$$

Step by step solution

01

Identify the substitution

In this exercise, we will use the trigonometric substitution \(x = 4\sin{\theta}\), since the integral has a form similar to \(\int\frac{1}{\sqrt{a^2 - x^2}}dx\) and \(16 = 4^2\). So, differentiate \(x\) with respect to \(\theta\) to get \(dx = 4\cos{\theta} d\theta\).
02

Replace x and dx with the substitution

Substitute \(x = 4\sin{\theta}\) and \(dx = 4\cos{\theta} d\theta\) in the given integral: $$\int\frac{4\cos{\theta} d\theta}{\sqrt{16 - (4\sin{\theta})^{2}}}$$
03

Simplify the integral

Simplify the integral by substituting and simplifying the terms inside the square root: $$\int\frac{4\cos{\theta} d\theta}{\sqrt{16 - 16\sin^2{\theta}}}$$ Now, recall the Pythagorean identity: \(1 - \sin^2{\theta} = \cos^2{\theta}\). The integral becomes: $$\int\frac{4\cos{\theta} d\theta}{\sqrt{16\cos^2{\theta}}}$$ Further, simplify the integral by taking the square root of \(16\cos^2{\theta}\): $$\int\frac{4\cos{\theta} d\theta}{4\cos{\theta}}$$
04

Evaluate the simplified integral

The integral is now simplified to: $$\int d\theta$$ Integrate with respect to \(\theta\) to get: $$\int d\theta = \theta + C$$ where \(C\) is the constant of integration.
05

Convert back to x

Now, we must convert back to the original variable \(x\). Recall the substitution we made, \(x = 4\sin{\theta}\). To find \(\theta\), we can rearrange this equation to find \(\sin{\theta}\): $$\sin{\theta} = \frac{x}{4}$$ Use the inverse sine function to find \(\theta\): $$\theta = \arcsin\left(\frac{x}{4}\right)$$ Substitute this back into the expression for the integral: $$\arcsin\left(\frac{x}{4}\right) + C$$ So, the evaluated integral is: $$\int \frac{dx}{\sqrt{16 - x^2}} = \arcsin\left(\frac{x}{4}\right) + C$$

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Key Concepts

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

Indefinite Integral
An indefinite integral represents a family of functions whose derivative yields the original function under the integral sign. It's also known as an antiderivative. The symbol \( \int \) is used to denote an integral, and the process of finding an integral is called integration. In our example, \( \int \frac{dx}{\sqrt{16 - x^2}} \) represents all functions whose derivative is the function \( \frac{1}{\sqrt{16 - x^2}} \) with respect to \( x \).

Performing integration requires recognizing patterns and often involves a range of techniques. The result includes a constant of integration, represented by 'C', since the derivative of a constant is zero. This constant must be added to account for all possible solutions that differ by a constant value.
Inverse Trigonometric Functions
Inverse trigonometric functions allow us to find the angle associated with a given trigonometric ratio. Commonly, these functions include \( \arcsin \), \( \arccos \), and \( \arctan \) among others, which are inverses to the sine, cosine, and tangent functions, respectively. In our integral example, we use the \( \arcsin \) function to find the angle \( \theta \) for a given sine value. When we solve for \( x \) as \( x = 4\sin{\theta} \) and then find \( \sin{\theta}=\frac{x}{4} \) to obtain \( \theta = \arcsin\left(\frac{x}{4}\right) \).

This process is crucial when applying trigonometric substitution in integrals as it allows us to return to the original variable after integration.
Pythagorean Identity
The Pythagorean identity in trigonometry relates the square of sine and cosine functions of the same angle and is derived from the Pythagorean theorem. This identity states that for any angle \( \theta \), the equation \( \sin^2{\theta} + \cos^2{\theta} = 1 \) holds true. It's a fundamental relation in trigonometry that allows the simplification of expressions involving \( \sin \) and \( \cos \) terms.

In the context of our integral example, this identity transformed \( \sqrt{16 - 16\sin^2{\theta}} \) into \( \sqrt{16\cos^2{\theta}} \) after substituting \( \sin^2{\theta} \) for \( 1 - \cos^2{\theta} \) which greatly simplified the integration process.
Integration Techniques
Integration techniques encompass a variety of methods used when finding the integral of a function is not straightforward. Common techniques include substitution, integration by parts, partial fractions, and trigonometric substitution, among others. Trigonometric substitution, as seen in our example, is particularly helpful when dealing with integrals containing square roots of quadratic expressions. Here, we chose the substitution \( x = 4\sin{\theta} \) because the expression within the square root resembled \( \sqrt{a^2 - x^2} \) which corresponds to a known Pythagorean identity.

The chosen substitution thereupon led to a straightforward integral that could be easily evaluated. Understanding when and how to apply these techniques is crucial for solving complex integrals that one may encounter in calculus.

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Most popular questions from this chapter

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t)\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume that \(s\) is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=e^{a t} \longrightarrow F(s)=\frac{1}{s-a}$$

\(A n\) integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or equivalently \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int_{0}^{\pi / 2} \frac{d \theta}{\cos \theta+\sin \theta}$$

Use symmetry to evaluate the following integrals. a. \(\int_{-\infty}^{\infty} e^{-|x|} d x\) b. \(\int_{-\infty}^{\infty} \frac{x^{3}}{1+x^{8}} d x\)

A remarkable integral It is a fact that \(\int_{0}^{\pi / 2} \frac{d x}{1+\tan ^{m} x}=\frac{\pi}{4}\) for all real numbers \(m .\) a. Graph the integrand for \(m=-2,-3 / 2,-1,-1 / 2,0,1 / 2\) \(1,3 / 2,\) and \(2,\) and explain geometrically how the area under the curve on the interval \([0, \pi / 2]\) remains constant as \(m\) varies. b. Use a computer algebra system to confirm that the integral is constant for all \(m.\)

Use the reduction formulas in a table of integrals to evaluate the following integrals. $$\int \tan ^{4} 3 y d y$$

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