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Chapter 6: Problem 52

Evaluate the integral. \(\int \frac{x}{\sqrt{1+x^{2}}} d x\)

Short Answer

Expert verified
\(\sqrt{1+x^2} + C\)

Step by step solution

01

Identify the Integration Technique

The integral \( \int \frac{x}{\sqrt{1+x^{2}}} \, dx \) can be solved using a substitution method. When you recognize the structure involving \(1+x^2\) under a square root, consider using trigonometric or hyperbolic substitution.
02

Choose a Substitution

For this integral, use the substitution \( x = \sinh(u) \). This choice simplifies \( \sqrt{1 + x^2} \) to \( \cosh(u) \). Also, compute \( dx = \cosh(u) \, du \).
03

Substitute and Simplify

Substitute \( x = \sinh(u) \) and \( dx = \cosh(u) \, du \) into the integral:\[\int \frac{\sinh(u)}{\cosh(u)} \cosh(u) \, du = \int \sinh(u) \, du\]This simplifies the integral as the \( \cosh(u) \) terms cancel.
04

Integrate with Respect to \( u \)

Now, integrate \( \int \sinh(u) \, du \). The integral of \( \sinh(u) \) is \( \cosh(u) + C \). So, we have:\[\cosh(u) + C\]
05

Substitute Back for \( x \)

Since \( u = \sinh^{-1}(x) \) was our substitution, and \( \cosh(u) = \sqrt{1+x^2} \), substitute back to get the integral in terms of \( x \):\[\sqrt{1+x^2} + C\]
06

Write the Final Answer

The evaluation of the integral is:\[\int \frac{x}{\sqrt{1+x^2}} \, dx = \sqrt{1+x^2} + C\]

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Key Concepts

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

Trigonometric Substitution
Trigonometric substitution is a useful technique in calculus for evaluating integrals involving roots of quadratic expressions. It's particularly helpful with expressions like \( \sqrt{a^2 + x^2} \), \( \sqrt{a^2 - x^2} \), and \( \sqrt{x^2 - a^2} \). In this technique, we substitute a trigonometric function for \( x \) to simplify the integral.
  • For \( \sqrt{1 + x^2} \), we can use either \( x = \tan(\theta) \) or a hyperbolic substitution as in the problem above, \( x = \sinh(u) \).
  • For \( \sqrt{a^2 - x^2} \), use \( x = a \sin(\theta) \).
  • For \( \sqrt{x^2 - a^2} \), substitute \( x = a \sec(\theta) \).
By making a trigonometric substitution, the integral becomes easier to manage as trigonometric functions have simpler derivatives and integrals. Thus, what might seem like a complex integral can often be transformed into a more standard form that we know how to solve.
Hyperbolic Substitution
Hyperbolic substitution is another substitution method used to solve integrals involving expressions similar to \( \sqrt{1 + x^2} \). Hyperbolic functions, like hyperbolic sine and cosine, share useful properties with their trigonometric counterparts, but are often more suited for certain integrals.
  • For expressing \( x \), we use \( x = \sinh(u) \), which simplifies \( \sqrt{1 + x^2} \) to \( \cosh(u) \).
  • Similarly, derivatives of hyperbolic functions are straightforward: \( \frac{d}{dx} \sinh(u) = \cosh(u) \), and \( \frac{d}{du} \cosh(u) = \sinh(u) \).
This problem is a perfect candidate for hyperbolic substitution as it quickly reduces the integral to a simpler form, as the \( \cosh(u) \) terms cancel, leaving us with an easy integral of \( \sinh(u) \). Such techniques can often avoid the presence of trigonometric functions altogether, simplifying integration when specific transformation patterns are recognized.
Indefinite Integrals
Indefinite integrals, denoted by the integral symbol \( \int \) followed by a function, represent the family of all antiderivatives of a function. They are termed 'indefinite' because they contain an arbitrary constant \( C \) rather than producing a specific value:
  • The fundamental theorem of calculus connects definite and indefinite integrals, stating that differentiation and integration are inverse processes.
  • When we calculate an indefinite integral, the result is a function plus a constant, \( C \), which represents the range of all possible vertical shifts of the antiderivative.
In our exercise, after applying hyperbolic substitution and integration, we are left with \( \cosh(u) + C \). When terms are transformed back, \( \sqrt{1 + x^2} + C \) appears as the solution, illustrating how indefinite integrals neatly encapsulate the "area under the curve" concept stretched over an indefinite, not bounded, interval.

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Most popular questions from this chapter

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. $$\int_{1}^{\infty} \frac{1}{x^{2}+x} d x$$

\(35-40=\) Make a substitution to express the integrand as a rational function and then evaluate the integral. $$\int \frac{e^{2 x}}{e^{2 x}+3 e^{x}+2} d x$$

(a) Find the approximations \(T_{8}\) and \(M_{8}\) for the integral \(\int_{0}^{1} \cos \left(x^{2}\right) d x\) (b) Estimate the errors in the approximations of part (a). (c) How large do we have to choose \(n\) so that the approximations \(T_{n}\) and \(M_{n}\) to the integral in part (a) are accurate to within 0.0001\(?\)

(a) If \(g(x)=1 /(\sqrt{x}-1),\) use your calculator or computer to make a table of approximate values of \(\int_{2}^{t} g(x) d x\) for \(t=5,10,100,1000,\) and \(10,000 .\) Does it appear that \(\int_{2}^{\infty} g(x) d x\) is convergent or divergent? (b) Use the Comparison Theorem with \(f(x)=1 / \sqrt{x}\) to show that \(\int_{2}^{\infty} g(x) d x\) is divergent. (c) Illustrate part (b) by graphing \(f\) and \(g\) on the same screen for 2\(\leqslant x \leqslant 20 .\) Use your graph to explain intuitively why \(\int_{2}^{\infty} g(x) d x\) is divergent.

Sketch the region and find its area (if the area is finite). $$S=\\{(x, y) |-2
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