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Chapter 8: Problem 53

Evaluate the integrals by making a substitution (possibly trigonometric) and then applying a reduction formula. $$\int_{0}^{1} 2 \sqrt{x^{2}+1} d x$$

Short Answer

Expert verified
The integral evaluates to \( \ln(1+\sqrt{2}) + 1 \).

Step by step solution

01

Identify the Substitution

We start by recognizing that the given integral involves the expression \( \sqrt{x^2 + 1} \). Typically, when we see \( x^2 + a^2 \) in an integral, a trigonometric substitution is useful. In this case, we use the substitution \( x = \sinh(u) \), so \( dx = \cosh(u) \, du \). This substitution simplifies \( \sqrt{x^2 + 1} \) to \( \cosh(u) \).
02

Change of Variables

Applying the substitution \( x = \sinh(u) \), we change the integral bounds: When \( x = 0 \), \( u = \sinh^{-1}(0) = 0 \); when \( x = 1 \), \( u = \sinh^{-1}(1) = \ln(1 + \sqrt{2}) \). The integral becomes \[ \int_{0}^{\ln(1+\sqrt{2})} 2 \cosh^2(u) du. \]
03

Simplify and Integrate Using Reduction Formula

Recall the identity \( \cosh^2(u) = \frac{1 + \cosh(2u)}{2} \). The integral becomes \[ \int_{0}^{\ln(1+\sqrt{2})} (1 + \cosh(2u)) du. \] Splitting the integral, we have \[ \int_{0}^{\ln(1+\sqrt{2})} du + \int_{0}^{\ln(1+\sqrt{2})} \cosh(2u) \, du. \] The first integral is simple: \( \ln(1+\sqrt{2}) \). For the second part, use the fact that \( \int \cosh(2u) \, du = \frac{1}{2}\sinh(2u) \).
04

Solve the Integrals

First integral gives \( \ln(1+\sqrt{2}) \). For the second, integrate \( \frac{1}{2} \sinh(2u) \) from 0 to \( \ln(1+\sqrt{2}) \). Plugging in the bounds, the second integral equals \( \sinh(\ln(1+\sqrt{2})). \)
05

Evaluate and Combine

Using \( \sinh(\ln(1+\sqrt{2})) = \frac{1}{2}((1+\sqrt{2})^2 - (1+\sqrt{2})^{-2}) \), simplify this expression. This yields \( \ln(1+\sqrt{2}) + 1 \) by combining terms and simplifying.

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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 branch of mathematics that deals with the concept of integrals. It is primarily used to find
  • areas under curves,
  • volumes of solids of revolution,
  • and other quantities that can be represented as limits of sums.
In solving the integral \[\int_{0}^{1} 2 \sqrt{x^{2}+1} \, dx, \] we work with definite integrals, focusing on calculating the area under the curve of the given function from one point to another. To approach this problem, trigonometric substitution is employed, which is a common technique in integral calculus. By using substitutions, complex expressions can be simplified into easier forms. Here, substituting \( x = \sinh(u) \) allows transforming the expression \( \sqrt{x^2 + 1} \) into \( \cosh(u) \), simplifying the integration process.
Reduction Formula
A reduction formula is a technique used in calculus to simplify the process of solving integrals, especially those that are recursive in nature.
Reduction formulas allow breaking down a complicated integral into simpler parts, often involving smaller or easier-to-evaluate integrals. In this example, the reduction formula used is derived from the hyperbolic identity \[\cosh^2(u) = \frac{1 + \cosh(2u)}{2}. \]This identity helps to separate the integral \[\int \cosh^2(u) \, du \]into two simpler integrals: \[\int du \] and \[\int \cosh(2u) \, du. \] The technique cleverly reduces the original integral into more manageable parts, solving them independently and then combining the results. The power of the reduction formula lies in turning what seems to be hard integrals into simple tasks, by identifying underlying patterns or relationships.
Hyperbolic Functions
Hyperbolic functions are analogs of the trigonometric functions but for the hyperbola, rather than the circle. They include functions like
  • sinh (hyperbolic sine),
  • cosh (hyperbolic cosine),
  • tanh, and more.
These functions have properties similar to trigonometric functions, but they exhibit exponential growth. For instance, the identities like \[\cosh^2(u) = \frac{1 + \cosh(2u)}{2} \] echos the trigonometric identity \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\).
In the substitution method used in the solution, hyperbolic functions help simplify the integral. By replacing \( x \) with \( \sinh(u) \), the integral tackles the naturally occurring \( \sqrt{x^2 + 1} \), which becomes \( \cosh(u) \). Interchanging between these forms makes the integration process easier and ties into their useful properties, notably when simplifying and evaluating integrals that could otherwise be rather complex using standard algebraic methods alone.

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Most popular questions from this chapter

Evaluate the integrals in Exercises \(39-50\). $$\begin{aligned} &\int \frac{1}{x\left(x^{4}+1\right)} d x\\\ &\left(\text {Hint: Multiply by } \frac{x^{3}}{x^{3}} .\right) \end{aligned}$$

Integration by parts leads to a rule for integrating es that usually gives good results:$$\begin{aligned}\int f^{-1}(x) d x &=\int y f^{\prime}(y) d y \\\&=y f(y)-\int f(y) d y \\ &=x f^{-1}(x)-\int f(y) d y\end{aligned}$$. The idea is to take the most complicated part of the integral, in this case \(f^{-1}(x)\), and simplify it first. For the integral of \(\ln x,\) we get $$\begin{aligned}\int \ln x \, d x &=\int y e^{y} d y \\\&=y e^{y}-e^{y}+C \\\&=x \ln x-x+C \end{aligned}$$ For the integral of \(\cos ^{-1} x\) we get $$\begin{aligned}\int \cos ^{-1} x d x &=x \cos ^{-1} x-\int \cos y \, d y \\\&=x \cos ^{-1} x-\sin y+C \\\&=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C\end{aligned}$$ Use the formula $$\int f^{-1}(x) d x=x f^{-1}(x)-\int f(y) d y$$ Express your answers in terms of \(x.\) $$\int \tan ^{-1} x d x$$

In Exercises \(9-16,\) express the integrand as a sum of partial fractions and evaluate the integrals. $$\int \frac{x+3}{2 x^{3}-8 x} d x$$

Many chemical reactions are the result of the interaction of two molecules that undergo a change to produce a new product. The rate of the reaction typically depends on the concentrations of the two kinds of molecules. If \(a\) is the amount of substance \(A\) and \(b\) is the amount of substance \(B\) at time \(t=0,\) and if \(x\) is the amount of product at time \(t,\) then the rate of formation of \(x\) may be given by the differential equation $$\frac{d x}{d t}=k(a-x)(b-x)$$ or $$\frac{1}{(a-x)(b-x)} \frac{d x}{d t}=k$$ where \(k\) is a constant for the reaction. Integrate both sides of this equation to obtain a relation between \(x\) and \(t\) (a) if \(a=b,\) and (b) if \(a \neq b .\) Assume in each case that \(x=0\) when \(t=0\)

In Exercises \(9-16,\) express the integrand as a sum of partial fractions and evaluate the integrals. $$\int \frac{2 x+1}{x^{2}-7 x+12} d x$$
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