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In Exercises 47-50, find the indefinite integrals, if possible, using the formulas and techniques you have studied so far in the text. $$\begin{array}{l}{\text { (a) } \int \frac{1}{\sqrt{1-x^{2}}} d x} \\ {\text { (b) } \int \frac{x}{\sqrt{1-x^{2}}} d x} \\ {\text { (c) } \int \frac{1}{x \sqrt{1-x^{2}}} d x}\end{array}$$

Short Answer

Expert verified
(a) \(\sin^{-1}(x) + C\), (b) \(-\sqrt{1-x^{2}} + C\), and (c) \(-\sqrt{2x-x^2} + \ln |1+x + \sqrt{2x-x^2}| + C\).

Step by step solution

01

Solving (a)

The integral provided is \(\int \frac{1}{\sqrt{1-x^{2}}} dx\). Use the trigonometric substitution, where \(x = \sin(\theta)\), also don't forget to change dx using the differential of sin, which is \(\cos(\theta)\). Therefore, the integral becomes \(\int d\theta\), which when evaluated gives \(\theta+C\). To substitute \(\theta\) back in terms of \(x\), take into account that \(\sin^{-1}(x) = \theta\). The final answer is \(\sin^{-1}(x) + C\).
02

Solving (b)

The integral provided is \(\int \frac{x}{\sqrt{1-x^{2}}} dx\). Again use the trigonometric substitution, where \(x = \sin(\theta)\). The dx will be \(\cos(\theta) d\theta\). Therefore, the integral becomes \(\int \sin(\theta) d\theta\), which when evaluated gives \(-\cos(\theta) + C\). To substitute \(\theta\) back in terms of \(x\), remember that \(\cos(\theta) = \sqrt{1 - x^2}\). The final result is \(-\sqrt{1-x^{2}} + C\).
03

Solving (c)

The integral provided is \(\int \frac{1}{x\sqrt{1-x^{2}}} dx\). This is a difficult one, but the trick will be to multiply and divide the integrand by \(1 + x\). Using the substitution \(u = 1+x\), dv is defined as \(-\frac{1}{x^2}dx\). Integration by parts can then be performed using \(u\) and \(dv\). After substituting \(x = u - 1\) and simplifying the integral, it is possible to solve the integral which results in \(-\sqrt{1-u^2} + \ln |u + \sqrt{1 - u^2}| + C\). Substituting \(u\) back in terms of \(x\) gives our final result of \(-\sqrt{2x-x^2} + \ln |1+x + \sqrt{2x-x^2}| + C\).

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Key Concepts

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

trigonometric substitution
Trigonometric substitution is a powerful technique used to simplify integrals involving expressions similar to \( \sqrt{a^2 - x^2} \), \( \sqrt{x^2 - a^2} \), or \( \sqrt{a^2 + x^2} \). This method involves replacing a variable with a trigonometric function to take advantage of Pythagorean identities, ultimately simplifying the integrand.

For example, to solve the integral \( \int \frac{1}{\sqrt{1-x^2}} \, dx \), we substitute \({x = \sin(\theta)}\). This substitution changes the integral into a form that is easier to manage: \({ \int d\theta }\), resulting in \({ \theta + C }\). Substituting back with \({ \theta = \sin^{-1}(x) }\) gives us the solution: \({ \sin^{-1}(x) + C }\).

It’s crucial to remember to convert the integrals back to the original variable after integrating, using trigonometric identities or inverse functions when necessary. This ensures the solution reflects the variable presented in the original problem.
integration by parts
Integration by parts is a technique derived from the product rule of differentiation. It is useful when integrating products of functions. The formula is given by:

\[ \int u \, dv = uv - \int v \, du \]

When solving integrals that appear complex due to products, selecting appropriate \( u \) and \( dv \) is key to simplifying the integral process.

In the exercise part (c), \( \int \frac{1}{x\sqrt{1-x^{2}}} \, dx \), we can cleverly tackle this by considering a transformation and then applying integration by parts. First, recognize the tricky parts of the integrand, then choose \( u = 1+x \) and \( dv = -\frac{1}{x^2} dx \). The integration by parts breaks down this complexity, leading to an integral that is more tractable.

Remember to keep track of your substitutions and adjustments carefully. Practicing this method with a variety of integral types helps reinforce understanding and increases proficiency.
integration techniques
Mastering integration techniques involves understanding several methods that help solve a wide array of integrals. Each technique has its scenarios where it shines, aiding in handling integrals that might otherwise seem too complex.

Some of the crucial integration techniques include:
  • **Basic Integration Rules:** Knowing the integral forms of basic functions and recognizing them in more complex integrals.
  • **Trigonometric Substitution:** Useful in transforming integrals involving \(\sqrt{1-x^2 }\)-like factors.
  • **Integration by Parts:** Useful for handling integrals involving products of functions.
  • **Partial Fraction Decomposition:** Handy for rational functions where the degree of the numerator is lower than that of the denominator.
Applied effectively, these techniques will allow you to assess and approach a complex integral strategically. Understanding which technique applies best to different integrals is a skill that develops with practice and experience. Each technique simplifies the integration process and expands your problem-solving toolkit.

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Most popular questions from this chapter

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