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There is a good chance that your CAS will not be able to evaluate these integrals as stated. If this is so, make a substitution that converts the integral into one that your CAS can evaluate. $$ \int(x \cos x+\sin x) \sqrt{1+x^{2} \sin ^{2} x} d x $$

Short Answer

Expert verified
The integral simplifies to \( \sinh^{-1}(x \sin x) + C \).

Step by step solution

01

Analyze the Integral

We begin by looking at the integral \( \int(x \cos x+\sin x) \sqrt{1+x^{2} \sin ^{2} x} \, dx \). It has a composite inner function in the square root making it difficult for direct evaluation.
02

Identify Substitution Candidates

Notice that the expression \( x \sin x \) appears inside the square root. We can let \( u = x \sin x \), as it simplifies the expression under the root structure.
03

Differentiate and Substitute

Differentiate \( u = x \sin x \) to find \( du \). Using the product rule, \( du = (x \cos x + \sin x) \, dx \). Notice that this matches part of our integrand, allowing a substitution.
04

Substitute into the Integral

Replace \( x \cos x + \sin x \, dx \) with \( du \) and \( x \sin x \) in the integrand, yielding \( \int \sqrt{1 + u^2} \, du \).
05

Evaluate the New Integral

Now, the integral \( \int \sqrt{1 + u^2} \, du \) is simpler. Many CAS systems can handle this integral directly, often using hyperbolic functions, recognized as the integral form related to \( \sinh^{-1}(u) \).
06

Re-substitute \( u \) Back to \( x \)

After solving, we translate back to the initial terms. As \( u = x \sin x \), express \( \sinh^{-1}(u) \) as \( \sinh^{-1}(x \sin x) \).
07

Finalize the Solution

Express the solution and apply the limits if it was a definite integral. The indefinite integral becomes \( \sinh^{-1}(x \sin x) + C \), where \( C \) is the constant of integration.

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Key Concepts

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

Substitution Method
The substitution method is a powerful integration technique used to simplify complex integrals, especially those involving composite functions. The goal is to transform the original integral into a more manageable form by introducing a new variable. This is often useful when your original expression includes cumbersome and convoluted terms, like nested functions or complex expressions under radicals.

In our example, the integral \( \int(x \cos x + \sin x) \sqrt{1 + x^{2} \sin^{2} x} \, dx \) was initially quite complex due to the square root containing the expression \( x^2 \sin^2 x \). By recognizing \( u = x \sin x \), we simplify the expression inside the square root to \( \sqrt{1 + u^2} \), making the integral much easier to evaluate.

When applying substitution:
  • Choose a substitution \( u = g(x) \) that simplifies the integral expression.
  • Differentiate \( u \) with respect to \( x \) to get \( du = g'(x) \, dx \).
  • Replace all \( x \)-terms and \( dx \) in the integral with \( u \) and \( du \).
  • Evaluate the new integral in terms of \( u \).
  • Re-substitute to express the final solution in terms of \( x \).
Product Rule
The product rule is a cornerstone differentiation technique used when taking derivatives of products of two functions. It is essential for correctly handling integrals that involve products when creating substitutions.

The product rule formula is given by:
\[\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)\]
This rule was applied in the step where we differentiated \( u = x \sin x \), resulting in:
\[du = (x \cos x + \sin x) \, dx\]
This corresponds to assigning \( f(x) = x \) and \( g(x) = \sin x \), thus making \( f'(x) = 1 \) and \( g'(x) = \cos x \). Substituting these into the product rule yields \( du \), which matched part of our original integrand.

Understanding the product rule is crucial because it allows us to successfully manipulate and simplify products of functions, especially when preparing for substitution during integration.
Hyperbolic Functions
Hyperbolic functions are analogs of trigonometric functions but for the hyperbola, similar in form yet different in properties. They play an important role in some integration scenarios, particularly when dealing with integrals like \( \int \sqrt{1 + u^2} \, du \).

The hyperbolic sine and cosine functions, denoted as \( \sinh(x) \) and \( \cosh(x) \), are defined by:
\[\sinh(x) = \frac{e^x - e^{-x}}{2}\]
\[\cosh(x) = \frac{e^x + e^{-x}}{2}\]
These functions are useful due to their relationships with the identities \( \cosh^2(x) - \sinh^2(x) = 1 \), which mirrors the Pythagorean identity in trigonometry.

When encountering an integral like \( \int \sqrt{1 + u^2} \, du \), hyperbolic functions often simplify the process. Specifically, this integral can be related to \( \sinh^{-1}(u) \), as many CAS or advanced integration techniques recognize it as a standard form.

Utilizing hyperbolic functions in integration allows for neat and elegant solutions to otherwise cumbersome integrals.

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Most popular questions from this chapter

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) or \(u=(x+a)^{1 / n}\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$ \int \frac{d x}{\sqrt{x}+\sqrt[3]{x}} $$

A transform is a formula that converts or "transforms" one function into another. Transforms are used in applications to convert a difficult problem into an easier problem whose solution can then be used to solve the original difficult problem. The Laplace transform of a function \(f(t)\), which plays an important role in the study of differential equations, is denoted by \(\mathscr{L}\\{f(t)\\}\) and is defined by $$\mathscr{L}\\{f(t)\\}=\int_{0}^{+\infty} e^{-s t} f(t) d t$$ In this formula \(s\) is treated as a constant in the integration process; thus, the Laplace transform has the effect of transforming \(f(t)\) into a function of \(s .\) Use this formula in these exercises. Show that (a) \(\mathscr{L}\\{1\\}=\frac{1}{s}, s>0\) (b) \(\mathscr{L}\left\\{e^{2 t}\right\\}=\frac{1}{s-2}, s>2\) (c) \(\mathscr{L}\\{\sin t\\}=\frac{1}{s^{2}+1}, s>0\) (d) \(\mathscr{L}\\{\cos t\\}=\frac{s}{s^{2}+1}, s>0\).

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) or \(u=(x+a)^{1 / n}\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$ \int x \sqrt{x-2} d x $$

Find a function \(g(x)\) of the form $$ g(x)=A x^{2}+B x+C $$ whose graph contains the points \((m-\Delta x, f(m-\Delta x))\), \((m, f(m))\), and \((m+\Delta x, f(m+\Delta x))\), for the given function \(f(x)\) and the given values of \(m\) and \(\Delta x .\) Then verify Formula (11): $$ \int_{m-\Delta x}^{m+\Delta x} g(x) d x=\frac{\Delta x}{3}\left[Y_{0}+4 Y_{1}+Y_{2}\right] $$ where \(Y_{0}=f(m-\Delta x), Y_{1}=f(m)\), and \(Y_{2}=f(m+\Delta x)\). $$ f(x)=\sin ^{2}(\pi x) ; m=\frac{1}{6}, \Delta x=\frac{1}{6} $$

Approximate the integral using Simpson's rule \(S_{10}\) and compare your answer to that produced by a calculating utility with a numerical integration capability. Express your answers to at least four decimal places. $$ \int_{-1}^{2} x \sqrt{2+x^{3}} d x $$

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