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Chapter 8: Problem 29

Evaluate the integrals by using a substitution prior to integration by parts. \(\int \sin (\ln x) d x\)

Short Answer

Expert verified
\( \int \sin(\ln x)\,dx = \frac{1}{2} x \sin(\ln x) + C \).

Step by step solution

01

Choose a substitution

To simplify the integration, first use the substitution method. Set \( u = \ln x \), which implies that \( x = e^u \) and \( dx = e^u \, du \). Substitute \( u \) into the integral, replacing \( \ln x \) with \( u \) and \( dx \) with \( e^u \, du \):\[\int \sin(u) e^u \, du\].
02

Setup integration by parts

Use the integration by parts formula, which is \( \int u \, dv = uv - \int v \, du \). Choose \( v = \sin(u) \) and \( dw = e^u \, du \), then differentiate and integrate respectively: \( dv = \cos(u) \, du \) and \( w = e^u \).
03

Apply integration by parts

Apply the integration by parts formula to the integral \( \int \sin(u) e^u \, du \):\[ \int \sin(u) e^u \, du = e^u (-\cos(u)) - \int -\cos(u) e^u \, du \]which simplifies to\[ -e^u \cos(u) + \int \cos(u) e^u \, du \].
04

Repeat integration by parts

To solve \( \int \cos(u) e^u \, du \), use integration by parts again, setting \( v = \cos(u) \) and \( dw = e^u \, du \): \( dv = -\sin(u) \, du \) and \( w = e^u \). Applying the integration by parts formula gives:\[ \int \cos(u) e^u \, du = e^u \cos(u) - \int -\sin(u) e^u \, du \]which simplifies to\[ e^u \cos(u) + \int \sin(u) e^u \, du \].
05

Solve for the original integral

Look at the equation derived from steps 3 and 4:\[ \int \sin(u) e^u \, du = -e^u \cos(u) + e^u \cos(u) + \int \sin(u) e^u \, du \]Realize that the \( \int \sin(u) e^u \, du \) term appears on both sides. Simplify by subtracting across:\[ 2 \int \sin(u) e^u \, du = e^u \sin(u) \]which reduces to\[ \int \sin(u) e^u \, du = \frac{1}{2} e^u \sin(u) \].
06

Substitute back to original variable

We now substitute back \( u = \ln x \) which implies \( e^u = x \), to find the integral in terms of \( x \):\[ \int \sin (\ln x) \, dx = \frac{1}{2} x \sin(\ln x) + C \],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.

Integration by Parts
Integration by parts is a fundamental technique in calculus used to integrate the product of two functions. It is based on the product rule for differentiation which states that the derivative of a product of two functions is not simply the product of their derivatives. Instead, it follows a specific rule: \[ \int u \, dv = uv - \int v \, du \]
  • "\(u\)" and "\(dv\)" are chosen based on the functions involved in the integral.
  • The "\(uv\)" term represents the product of "\(u\)" and the integral of "\(dv\)," noted as "\(v\)."
  • The new integral, "\(\int v \, du\)," often tends to be simpler than the original.
This method is particularly useful when integrating the product of functions where standard integration techniques do not apply. Make sure to select "\(u\)" and "\(dv\)" wisely to make the integral more manageable.
Substitution Method
The substitution method, also known as integration by substitution, is a technique that simplifies an integral by changing its variable. This method is akin to reverse chain rule application. A substitution is chosen to transform the integral into a more straightforward form:
  • Identify a portion of the integral that can be substituted with a single variable, "\(u\)."
  • Find the derivative of this substitution to replace "\(dx\)" with "\(du\)."
  • Rewrite the integral using the new variable "\(u\)," making it easier to solve.
In this exercise, substituting \( u = \ln x \) converts the integral into a form easily solved using integration by parts. This substitution simplifies function components and aligns them effectively for subsequent steps.
Definite Integrals
Unlike indefinite integrals, definite integrals calculate the total accumulation of a quantity, often represented as the area under a curve, between two limits. Though this problem involves an indefinite integral, it's crucial to understand definite integrals since they contextualize integration results. The formula is:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]
  • \(f(x)\) is the function being integrated, and \(a\) and \(b\) are the limits.
  • \(F(x)\) is the antiderivative or integral of \(f(x)\).
  • The result represents the net area between \(a\) and \(b\) under the curve \(f(x)\).
Definite integrals factor into applications like physics and engineering, emphasizing their importance beyond finding antiderivatives.
Indefinite Integrals
An indefinite integral, unlike its definite counterpart, does not have upper or lower limits and thus lacks a specific numerical value. Instead, it represents a family of functions differentiated by a constant. The general form is:\[ \int f(x) \, dx = F(x) + C \]
  • "\(f(x)\)" is the function being integrated.
  • "\(F(x)\)" is its antiderivative.
  • "\(C\)" represents the constant of integration, accounting for any vertical shifts in antiderivative graphs.
In the given exercise, the result is an indefinite integral, \( \frac{1}{2} x \sin(\ln x) + C \), reflecting the general solution.Students must understand this concept because antiderivatives form the basis for applications across various fields like physics and economics.

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Most popular questions from this chapter

In Exercises \(29-36\) , use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. $$ \int \frac{d x}{x \sqrt{x^{2}-1}} $$

The infinite paint can or Gabriel's horn As Example 3 shows, the integral \(\int_{1}^{\infty}(d x / x)\) diverges. This means that the integral $$\int_{1}^{\infty} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x$$ which measures the surface area of the solid of revolution traced out by revolving the curve \(y=1 / x, 1 \leq x,\) about the \(x\) -axis, di- verges also. By comparing the two integrals, we see that, for every finite value \(b>1\) , $$\int_{1}^{b} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x>2 \pi \int_{1}^{b} \frac{1}{x} d x$$ (GRAPH NOT COPY) However, the integral $$\int_{1}^{\infty} \pi\left(\frac{1}{x}\right)^{2} d x$$ for the volume of the solid converges. (a) Calculate it. ( \(\mathbf{b} )\) This solid of revolution is sometimes described as a can that does not hold enough paint to cover its own interior. Think about that for a moment. It is common sense that a finite amount of paint cannot cover an infinite surface. But if we fill the horn with paint (a finite amount), then we will have covered an infinite surface. Explain the apparent contradiction.

Use a CAS to perform the integrations. Evaluate the integrals a. \(\int x \ln x d x \) b. \(\int x^{2} \ln x d x\) c. \(\int x^{3} \ln x d x\) d. What pattern do you see? Predict the formula for \(\int x^{4} \ln x d x\) and then see if you are correct by evaluating it with a CAS. e. What is the formula for \(\int x^{n} \ln x d x, n \geq 1 ?\) Check your answer using a CAS.

Find the length of the curve \(y=x^{2}\) , \(0 \leq x \leq \sqrt{3} / 2\).

a. Show that if \(f\) is even and the necessary integrals exist, then $$\int_{-\infty}^{\infty} f(x) d x=2 \int_{0}^{\infty} f(x) d x$$ b. Show that if \(f\) is odd and the necessary integrals exist, then $$\int_{-\infty}^{\infty} f(x) d x=0$$
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