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Evaluate the following integrals. $$\int \frac{\sqrt{16-x^{2}}}{x^{2}} d x$$

Short Answer

Expert verified
Answer: The result of evaluating the integral \(\int \frac{\sqrt{16-x^{2}}}{x^{2}} d x\) is \(\frac{64}{x} - 4x + C\), where C is the constant of integration.

Step by step solution

01

Substitute a trigonometric expression for the variable x

In order to simplify the integrand, substitute x with \(4\sin(\theta)\). This will help to simplify the square root in the denominator. $$x = 4\sin(\theta)$$ Now we find the differential. $$d x = \frac{d(4\sin(\theta))}{d \theta} d \theta = 4\cos(\theta) d \theta$$
02

Rewrite the integrand in terms of \(\theta\)

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

Simplify the integrand

Simplify the expression inside the square root and cancel out the common terms: $$\int \frac{\sqrt{16-16\sin^2(\theta)}}{(4\sin(\theta))^{2}} (4\cos(\theta) d \theta) = \int \frac{\sqrt{16(1-\sin^2(\theta))}}{16\sin^2(\theta)} (4\cos(\theta) d \theta)$$ Using the trigonometric identity \(1-\sin^2(\theta) = \cos^2(\theta)\): $$\int \frac{\sqrt{16\cos^2(\theta)}}{16\sin^2(\theta)}(4\cos(\theta) d \theta) = \int \frac{4\cos(\theta)}{\sin^2(\theta)}(4\cos(\theta) d \theta)$$
04

Perform the integration

Now integrate the simplified integrand with respect to \(\theta\): $$\int \frac{4\cos(\theta)}{\sin^2(\theta)}(4\cos(\theta) d \theta) = 16 \int \frac{\cos^2(\theta)}{\sin^2(\theta)} d\theta$$ Use the substitution \(u = \sin(\theta), \mathrm{d}u = \cos(\theta) \mathrm{d}\theta\): $$16 \int \frac{1-u^2}{u^2} d u = 16 \int \frac{1}{u^2} d u - 16 \int d u$$ Now, compute the integrals: $$16 \int \frac{1}{u^2} d u - 16 \int d u = -16 \int u^{-2} d u - 16 \int 1 d u = -16(-\frac{1}{u}) - 16(u) + C$$ Replace \(u\) with \(\sin(\theta)\): $$-16(-\frac{1}{\sin(\theta)}) - 16(\sin(\theta)) + C = \frac{16}{\sin(\theta)} - 16\sin(\theta) + C$$
05

Substitute back x

Now substitute \(\theta\) back to \(x\). Recall that \(x=4\sin(\theta)\), which means \(\sin(\theta)=\frac{x}{4}\). $$\frac{16}{\sin(\theta)} - 16\sin(\theta) + C = \frac{16}{\frac{x}{4}} - 16\left(\frac{x}{4}\right) + C = \frac{64}{x} - 4x + C$$ So the final result is: $$\int \frac{\sqrt{16-x^{2}}}{x^{2}} d x = \frac{64}{x} - 4x + C$$

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Key Concepts

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

Integration Techniques
Integration techniques can be seen as a set of tools or methods that help simplify complex integrals. When faced with integrals like \( \int \frac{\sqrt{16-x^{2}}}{x^{2}} \, dx \), techniques such as trigonometric substitution are invaluable.
Trigonometric substitution is often employed when the expression inside an integral involves expressions like \(a^2 - x^2\). To simplify this type of problem, we can substitute \(x\) with a trigonometric identity that relates to the form \(a\sin(\theta)\), because \(\sin^2(\theta) + \cos^2(\theta) = 1\). This transforms our original problem into a trigonometric problem which is generally easier to solve.
  • For example, we use \(x = 4\sin(\theta)\) to change the variable in our integral. This substitution simplifies the integral by using the identity \(16 - (4\sin(\theta))^2 = 16\cos^2(\theta)\).
  • After rewriting in terms of \(\theta\) and simplifying, you can then use basic integration techniques to solve the integral.
Applying these substitution methods greatly aids in breaking down complex integrals into more manageable forms.
Integrals Evaluation
Evaluating integrals, especially those that involve substitution techniques, requires a structured approach. Let's break down the process using the given integral problem.
The process begins with substituting a variable with a trigonometric expression to simplify the integrand. For example, substituting \(x = 4\sin(\theta)\) supports reducing complex radicals.
  • First, rewrite the entire integral in terms of \(\theta\). This includes both the variable and the differential \(dx\) changing to \(4\cos(\theta) d\theta\).
  • Once rewritten, simplify the resulting expression. Here, trigonometric identities help eliminate complex square roots or other difficult terms.
  • After simplification, integrate the simpler function obtained. This usually involves standard integrals known from basic calculus.
  • In this exercise: \[\int \frac{4\cos(\theta)}{\sin^2(\theta)}(4\cos(\theta) d\theta) \] becomes easier to handle and involves breaking it into components that can be individually integrated.
Finally, don't forget to substitute back the original variable to complete the evaluation. This ensures that your answer fits the format of the initial problem.
Trigonometric Identities
Trigonometric identities are essential tools in simplifying integral problems, especially when employing techniques like trigonometric substitution.
These identities help convert complex trigonometric expressions into simpler, more recognizable forms.
  • The fundamental identity \(\sin^2(\theta) + \cos^2(\theta) = 1\), used frequently in transformations, retains its utility here.
  • Another identity: \(1 - \sin^2(\theta) = \cos^2(\theta)\), commonly transforms squared sine functions into cosine and vice versa. It's incredibly useful when simplifying expressions under square roots.
  • When evaluating integrals, identities transform integrated terms into forms that standard calculus functions can handle more readily.
In this particular exercise, the identity \(1 - \sin^2(\theta) = \cos^2(\theta)\) was crucial. It allowed the simplification of the expression under the square root, moving from complex radicals to pure powers of cosines. With these identities, the challenge of complicated trigonometric integrands is lessened substantially, paving the way for a straightforward integration process.

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Most popular questions from this chapter

Use numerical methods or a calculator to approximate the following integrals as closely as possible. The exact value of each integral is given. $$\int_{0}^{\pi / 2} \ln (\sin x) d x=\int_{0}^{\pi / 2} \ln (\cos x) d x=-\frac{\pi \ln 2}{2}$$

Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals. $$\int \frac{\left(e^{3 x}+e^{2 x}+e^{x}\right)}{\left(e^{2 x}+1\right)^{2}} d x$$

It can be shown that $$\begin{array}{l}\int_{0}^{\pi / 2} \sin ^{n} x d x=\int_{0}^{\pi / 2} \cos ^{n} x d x= \\\\\quad\left\\{\begin{array}{ll}\frac{1 \cdot 3 \cdot 5 \cdot \cdots(n-1)}{2 \cdot 4 \cdot 6 \cdots n} \cdot \frac{\pi}{2} & \text { if } n \geq 2 \text { is an eveninteger } \\\\\frac{2 \cdot 4 \cdot 6 \cdots(n-1)}{3 \cdot 5 \cdot 7 \cdots n} & \text { if } n \geq 3 \text { is an odd integer. }\end{array}\right.\end{array}$$ a. Use a computer algebra system to confirm this result for \(n=2,3,4,\) and 5 b. Evaluate the integrals with \(n=10\) and confirm the result. c. Using graphing and/or symbolic computation, determine whether the values of the integrals increase or decrease as \(n\) increases.

The gamma function is defined by \(\Gamma(p)=\int_{0}^{\infty} x^{p-1} e^{-x} d x,\) for \(p\) not equal to zero or a negative integer. a. Use the reduction formula $$\int_{0}^{\infty} x^{p} e^{-x} d x=p \int_{0}^{\infty} x^{p-1} e^{-x} d x \quad \text { for } p=1,2,3, \ldots$$ to show that \(\Gamma(p+1)=p !(p\) factorial). b. Use the substitution \(x=u^{2}\) and the fact that \(\int_{0}^{\infty} e^{-u^{2}} d u=\frac{\sqrt{\pi}}{2}\) to show that \(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\).

Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x\). b. Change variables by letting \(y=1 / x\). c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) (and that both integrals converge). Conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x=0\). d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\). (Source: Mathematics Magazine \(59,1,\) Feb 1986 )

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