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Evaluate the integrals by making the indicated substitutions. (a) \(\int 2 x\left(x^{2}+1\right)^{23} d x ; u=x^{2}+1\) (b) \(\int \cos ^{3} x \sin x d x ; u=\cos x\) (c) \(\int \frac{1}{\sqrt{x}} \sin \sqrt{x} d x ; u=\sqrt{x}\) (d) \(\int \frac{3 x d x}{\sqrt{4 x^{2}+5}} ; u=4 x^{2}+5\) (e) \(\int \frac{x^{2}}{x^{3}-4} d x ; u=x^{3}-4\)

Short Answer

Expert verified
(a) \(\frac{(x^2+1)^{24}}{24} + C\); (b) \(-\frac{\cos^4 x}{4} + C\); (c) \(-2 \cos \sqrt{x} + C\); (d) \(\frac{3}{4} \sqrt{4x^2+5} + C\); (e) \(\frac{1}{3} \ln |x^3-4| + C\).

Step by step solution

01

Substitute for Integration (a)

The integral is \( \int 2x (x^2 + 1)^{23} \, dx \). To use the substitution \( u = x^2 + 1 \), differentiate to get \( du = 2x \, dx \). Therefore, \( 2x \, dx = du \). The integral becomes \( \int u^{23} \, du \).
02

Integrate with Respect to u (a)

Now, integrate \( \int u^{23} \, du \), which results in \( \frac{u^{24}}{24} + C \).
03

Substitute Back in Terms of x (a)

Since \( u = x^2 + 1 \), substitute back to get \( \frac{(x^2 + 1)^{24}}{24} + C \).
04

Substitute for Integration (b)

The integral is \( \int \cos^3 x \sin x \, dx \). Use the substitution \( u = \cos x \), so \( du = -\sin x \, dx \) or \( -du = \sin x \, dx \). The integral becomes \( -\int u^3 \, du \).
05

Integrate with Respect to u (b)

Now, integrate \( -\int u^3 \, du \), which results in \( -\frac{u^4}{4} + C \).
06

Substitute Back in Terms of x (b)

Substitute back using \( u = \cos x \), giving \( -\frac{\cos^4 x}{4} + C \).
07

Substitute for Integration (c)

The integral is \( \int \frac{1}{\sqrt{x}} \sin \sqrt{x} \, dx \). Use the substitution \( u = \sqrt{x} \), so \( x = u^2 \), \( dx = 2u \, du \). Thus, the integral becomes \( \int \sin u \cdot 2 \, du \) or \( 2 \int \sin u \, du \).
08

Integrate with Respect to u (c)

The integral \( 2 \int \sin u \, du \) becomes \( -2 \cos u + C \).
09

Substitute Back in Terms of x (c)

Substitute back using \( u = \sqrt{x} \), giving \( -2 \cos \sqrt{x} + C \).
10

Substitute for Integration (d)

The integral is \( \int \frac{3x}{\sqrt{4x^2 + 5}} \, dx \). Use the substitution \( u = 4x^2 + 5 \), so \( du = 8x \, dx \) or \( \frac{1}{8} \, du = x \, dx \). This makes the integral \( \frac{3}{8} \int \frac{1}{\sqrt{u}} \, du \).
11

Integrate with Respect to u (d)

The integral \( \frac{3}{8} \int u^{-1/2} \, du \) results in \( \frac{3}{8} \cdot \frac{u^{1/2}}{1/2} + C \), which simplifies to \( \frac{3}{4} \sqrt{u} + C \).
12

Substitute Back in Terms of x (d)

Substitute back using \( u = 4x^2 + 5 \), giving \( \frac{3}{4} \sqrt{4x^2 + 5} + C \).
13

Substitute for Integration (e)

The integral is \( \int \frac{x^2}{x^3 - 4} \, dx \). Use the substitution \( u = x^3 - 4 \), so \( du = 3x^2 \, dx \) or \( \frac{1}{3} \, du = x^2 \, dx \). This makes the integral \( \frac{1}{3} \int \frac{1}{u} \, du \).
14

Integrate with Respect to u (e)

The integral \( \frac{1}{3} \int \frac{1}{u} \, du \) results in \( \frac{1}{3} \ln |u| + C \).
15

Substitute Back in Terms of x (e)

Substitute back using \( u = x^3 - 4 \), giving \( \frac{1}{3} \ln |x^3 - 4| + C \).

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Key Concepts

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

Definite Integrals
Definite integrals are a fundamental concept in calculus, representing the signed area under a curve over a specific interval. Unlike indefinite integrals, which result in a family of functions, definite integrals yield a numerical value. This process involves:
  • Choosing the limits of integration to specify the interval.
  • Computing the antiderivative.
  • Using the Fundamental Theorem of Calculus to evaluate at the upper and lower limits.
Imagine you have a function representing speed over time. The definite integral of this function over a time interval gives you the total distance traveled. Remember that definite integrals can also account for areas below the x-axis, which are considered negative and can influence the final result. They are incredibly useful in real-world applications, such as finding distances, areas, and volumes.
Indefinite Integrals
Indefinite integrals, also known as antiderivatives, represent a set of all possible functions whose derivative is the original function you started with. Essentially, they do not have specific limits of integration—they are generalized, inclusive of a constant, designated as '+ C'. This constant accounts for all possible vertical shifts of the antiderivative graph. To compute an indefinite integral, consider these steps:
  • Identify the function you wish to integrate.
  • Find the antiderivative.
  • Add the constant of integration \( C \).
For example, if given \( f(x) = 2x \), an indefinite integral could be \( x^2 + C \). Each function within this family shares the same slope of \( f(x) \). Indefinite integrals form the basis for solving differential equations and finding functions given their derivatives.
U-Substitution
U-substitution, or integration by substitution, is an integration technique that simplifies integrals by making a substitution for a new variable \( u \). This method resembles the reverse of the chain rule in differentiation. To perform u-substitution effectively:
  • Select an expression inside the integral to replace with \( u \).
  • Differentiate this expression to find \( du \).
  • Rewrite the integral in terms of \( u \) and \( du \).
  • Integrate and convert back into the original variable \( x \).
This technique can turn complicated integrals into simpler ones. It is crucial to pick the substitution correctly so that \( du \) neatly replaces the differential part \( dx \). U-substitution is particularly beneficial when integrating powers of binomials or trigonometric expressions that seem too complex to handle directly.
Integration Techniques
Integration techniques encompass a variety of methods to tackle complex integrals. Each technique is suited for different types of integrands and scenarios. Common techniques include:
  • Basic integration rules for polynomials and exponents.
  • U-substitution for expressions requiring variable change.
  • Integration by parts for products of functions, derived from the product rule for differentiation.
  • Partial fraction decomposition for rational functions with polynomial denominators.
  • Trigonometric identities to simplify integrals involving sine, cosine, and other trigonometric functions.
Understanding when and how to apply these techniques is vital. For instance, u-substitution is ideal when an integral is the derivative of a simple function that can be isolated by a substitution. Mastery of these techniques opens up a wide range of possibilities for solving not just basic integrals but also more complex problems across mathematics and applied sciences.

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