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Chapter 5: Problem 9

Evaluate the indefinite integrals in Exercises \(1-12\) by using the given substitutions to reduce the integrals to standard form. $$ \int \sqrt{x} \sin ^{2}\left(x^{3 / 2}-1\right) d x, \quad u=x^{3 / 2}-1 $$

Short Answer

Expert verified
\( \frac{1}{3} \left[ (x^{3/2} - 1) - \frac{1}{2} \sin(2(x^{3/2} - 1)) \right] + C \).

Step by step solution

01

Substitution

First, identify the appropriate substitution for the problem. Given that \(u = x^{3/2} - 1\), we differentiate \(u\) with respect to \(x\): \( \frac{du}{dx} = \frac{3}{2} x^{1/2} \). Thus, \( du = \frac{3}{2} x^{1/2} \, dx \).
02

Solve for dx

Solve the expression for \(dx\) in terms of \(du\) and \(x\): - Rearrange the equation from Step 1 to find \(dx\): \( dx = \frac{2}{3} \frac{du}{x^{1/2}} \).
03

Substitute in the Integral

Substitute \(u\) and the expression for \(dx\) back into the integral: \[ \int \sqrt{x} \sin^2(u) \left(\frac{2}{3} \frac{du}{x^{1/2}}\right) = \int \frac{2}{3} \sin^2(u) \, du \].
04

Integrate with respect to u

The integral now has a standard form \( \int \sin^2(u) \, du \). Use the trigonometric identity for \( \sin^2(u) \):\( \sin^2(u) = \frac{1 - \cos(2u)}{2} \).Substitute this identity into the integral:\[ \frac{2}{3} \int \frac{1 - \cos(2u)}{2} \, du = \frac{1}{3} \int (1 - \cos(2u)) \, du \].The integral becomes:\[ \frac{1}{3} \left[ u - \frac{1}{2} \sin(2u) \right] + C \].
05

Back-substitute x

Re-substitute \(u = x^{3/2} - 1\) back into the result:\[ \frac{1}{3} \left[ (x^{3/2} - 1) - \frac{1}{2} \sin(2(x^{3/2} - 1)) \right] + C \].

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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 tool for solving integrals. It involves changing variables to simplify the integral into a more recognizable form. This is especially useful when the original integral is complex or has a composition of functions. In our exercise, we have the function \(u = x^{3/2} - 1\). By substituting \(u\) into the integral, we aim to simplify it.
The key steps include:
  • Identifying a substitution: Often, this is a part of the integrand that can be rearranged into a simpler form.
  • Differentiating the substitution function to find \( du \): This step helps us rewrite the original \( dx \) in terms of the new variable \( u \).
  • Rewriting the integral: Substitute \( u \) and \( du \) back into the integral, transforming it into a simpler problem.
By switching from \( x \) to \( u \), you often end up with an integral that is easier to manage, like the standard forms found in the integration tables. Always remember to convert back to the original variable at the end.
Trigonometric Identities
Trigonometric identities are essential tools for simplifying expressions and solving integrals that contain trigonometric functions. When dealing with integrals involving \( \sin^2, \, \cos^2 \,\) or other trigonometric powers, identities can help rewrite these functions into a form suitable for integration.
In this exercise, we use the identity for \( \sin^2(u) \):\[ \sin^2(u) = \frac{1 - \cos(2u)}{2} \]
  • This identity transforms the squared sine function into a combination of a constant and a cosine function, which are easier to integrate.
  • Applying these identities simplifies the integration process and turns a potentially complex integral into a straightforward one.
When you encounter powers of trigonometric functions within an integral, always consider using identities to simplify the process. This simplification is crucial for straightforward and effective integration.
Integration Techniques
Solving integrals often requires a variety of techniques, each suited to different types of integrals. Mastering these techniques increases your ability to handle a range of calculus problems. The discussed exercise leverages both substitution and trigonometric identities, but the broader world of integration includes even more techniques.
Common integration techniques include:
  • Substitution: Simplifies the integral by changing variables, as seen in our exercise.
  • Integration by Parts: Used when the product of two functions is involved. This method is based on the product rule for differentiation.
  • Trigonometric Integration: Uses trigonometric identities to simplify integrals containing trigonometric functions.
  • Partial Fraction Decomposition: Helpful for rational functions, breaking them into simpler fractions for integration.
Each technique requires practice to identify when and how to use them effectively. In our exercise, combining substitution with trigonometric identities showcases the versatility and necessity of understanding multiple strategies when working with indefinite integrals.

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Most popular questions from this chapter

In Exercises \(75-78,\) let \(F(x)=\int_{a}^{u(x)} f(t) d t\) for the specified \(a, u,\) and \(f\) . Use a CAS to perform the following steps and answer the questions posed. a. Find the domain of \(F\) b. Calculate \(F^{\prime}(x)\) and determine its zeros. For what points in its domain is \(F\) increasing? decreasing? c. Calculate \(F^{\prime \prime}(x)\) and determine its zero. Identify the local extrema and the points of inflection of \(F\) . d. Using the information from parts (a)-(c), draw a rough hand-sketch of \(y=F(x)\) over its domain. Then graph \(F(x)\) on your CAS to support your sketch. $$ a=0, \quad u(x)=1-x^{2}, \quad f(x)=x^{2}-2 x-3 $$

In Exercises \(75-78,\) let \(F(x)=\int_{a}^{u(x)} f(t) d t\) for the specified \(a, u,\) and \(f\) . Use a CAS to perform the following steps and answer the questions posed. a. Find the domain of \(F\) b. Calculate \(F^{\prime}(x)\) and determine its zeros. For what points in its domain is \(F\) increasing? decreasing? c. Calculate \(F^{\prime \prime}(x)\) and determine its zero. Identify the local extrema and the points of inflection of \(F\) . d. Using the information from parts (a)-(c), draw a rough hand-sketch of \(y=F(x)\) over its domain. Then graph \(F(x)\) on your CAS to support your sketch. $$ a=0, \quad u(x)=x^{2}, \quad f(x)=\sqrt{1-x^{2}} $$

Solve the initial value problems in Exercises \(53-58\). $$ \frac{d r}{d \theta}=3 \cos ^{2}\left(\frac{\pi}{4}-\theta\right), \quad r(0)=\frac{\pi}{8} $$

True, sometimes true, or never true? The area of the region between the graphs of the continuous functions \(y=f(x)\) and \(y=g(x)\) and the vertical lines \(x=a\) and \(x=b(a

Find the areas of the regions enclosed by the lines and curves in Exercises \(41-50 .\) $$ y=x^{2}-2 x \quad \text { and } \quad y=x $$
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