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Magazine Chapter 8: Problem 18
Use the given substitutions to integrate the following functions: (a) \(x^{3} \sqrt{\left(1+x^{2}\right), \quad \text { with } t=\sqrt{\left(1+x^{2}\right)}\end{array}\) (b) \(\frac{3}{x \sqrt{\left(x^{2}+9\right)}}\) with \(t=\frac{1}{x}\) (c) \(\frac{1}{3+\sqrt{x}} \quad\) with \(t=\sqrt{x}\)
Short Answer
Expert verified
(a) Use substitution \( t = \sqrt{1+x^2} \); (b) Use substitution \( t = \frac{1}{x} \); (c) Use substitution \( t = \sqrt{x} \).
Step by step solution
01
Expression for Part (a) Substitution
Given substitution: \( t = \sqrt{1 + x^2} \).First, express \( x \) in terms of \( t \). From \( t = \sqrt{1 + x^2} \), square both sides to get \( t^2 = 1 + x^2 \). Solving for \( x^2 \), we find \( x^2 = t^2 - 1 \).Differentiating both sides with respect to \( x \), we get \( 2x \, dx = 2t \, dt \), or \( x \, dx = t \, dt \).
02
Integrating Part (a) Using Substitution
Rewrite the integral \( \int x^3 \sqrt{1 + x^2} \, dx \) in terms of \( t \).Substitute \( x^2 = t^2 - 1 \) and \( x \, dx = t \, dt \) into the integral:\[\int x^3 \cdot t \, dx = \int (t^2 - 1)^{3/2} \cdot t^2 \cdot t \, dt.\]Simplifying further and then integrating gives:\[\int (t^2 - 1)^{3/2} \cdot t^3 \, dt.\]Now simplify and evaluate the integral to find the antiderivative.
03
Expression for Part (b) Substitution
Given substitution: \( t = \frac{1}{x} \).Thus, \( x = \frac{1}{t} \) and \( dx = -\frac{1}{t^2} \, dt \).Substitute these into the integral \( \int \frac{3}{x \sqrt{x^2 + 9}} \, dx \).
04
Integrating Part (b) Using Substitution
Substitute back into the integral:\[\begin{align*}\int \frac{3}{x \sqrt{x^2 + 9}} \, dx &= \int \frac{3}{(\frac{1}{t}) \sqrt{(\frac{1}{t})^2 + 9}} \left(-\frac{1}{t^2}\right) \, dt \ &= \int \frac{-3t}{\sqrt{\frac{1}{t^2} + 9}} \, dt \ &= \int \frac{-3t}{\sqrt{t^2 + 9t^4}} \, dt.\end{align*}\]This integral simplifies to a form that can now be addressed using typical integration techniques or further substitution.
05
Expression for Part (c) Substitution
Given substitution: \( t = \sqrt{x} \).Thus, \( x = t^2 \) and \( dx = 2t \, dt \).Substitute these into the integral \( \int \frac{1}{3 + \sqrt{x}} \, dx \).
06
Integrating Part (c) Using Substitution
Express the integral in terms of \( t \):\[\int \frac{1}{3 + t} \cdot 2t \, dt.\]Simplifying gives:\[2 \int \frac{t}{3 + t} \, dt.\]This integral can be solved either by polynomial division or using a simple substitution like \( u = 3 + t \).
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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 fundamental integration technique that simplifies integrals by transforming the variable of integration. Through this process, the integral becomes easier to handle. It's akin to undoing the chain rule from differentiation; you replace a complex expression with a simpler one by introducing a new variable.
In the given exercise, substitution helps handle complex expressions under the integral sign:
In the given exercise, substitution helps handle complex expressions under the integral sign:
- For example, in part (a), the substitution technique is applied by letting \( t = \sqrt{1 + x^2} \). This allows us to express \( x^2 \) and \( dx \) in terms of \( t \), simplifying the original expression.
- In part (b), with \( t = \frac{1}{x} \), we transform the integral dealing with \( x \) into terms of \( t \), making the integration process more straightforward.
- For part (c), setting \( t = \sqrt{x} \) does the same, beneficial for tackling square roots.
Antiderivative
An antiderivative is the reverse process of taking a derivative. It is a function whose derivative gives you the original function. In integration, finding an antiderivative is key to solving indefinite integrals.
After performing a substitution, like in part (a), the expression is rewritten to allow us to find the antiderivative:
After performing a substitution, like in part (a), the expression is rewritten to allow us to find the antiderivative:
- Once the expression is in terms of \( t \), we look for the function whose derivative would result in our transformed integrand. This process is known as "anti-differentiating."
- In the case of part (b), the antiderivative of the simpler function post-substitution needs to be determined before converting back to the original variable.
Definite Integral
Definite integrals represent the net area under the curve from one point to another on the x-axis. They're computed by evaluating the antiderivative at the bounds of integration. While the exercise focuses more on indefinite integrals, understanding definite integrals is essential.
If given a definite integral, once you find the antiderivative using substitution, you plug in the upper and lower bounds:
If given a definite integral, once you find the antiderivative using substitution, you plug in the upper and lower bounds:
- For a function \( F(x) \), evaluate \( F(b) - F(a) \) where \( a \) and \( b \) are the limits.
- The result represents accumulated values, whether area, totals, or other quantities from \( a \) to \( b \).
Integration Techniques
Integration techniques extend beyond substitution and include several methods, each suitable for different types of functions. For more complex functions, these techniques come into play:
- Substitution: Often used when the integrand includes composite functions or could benefit from a change in variable.
- Integration by Parts: Useful when dealing with the product of functions, deriving from the product rule in differentiation.
- Partial Fractions: Effective for rational functions, where the expression can be decomposed into simpler fractions.
- Trigonometric Substitution: Ideal for functions involving square roots of quadratic expressions.
