Problem 48 If appropriate, evaluate the fol... [FREE SOLUTION]

Chapter 6: Problem 48

If appropriate, evaluate the following integrals by substitution. If substitution is not appropriate, say so, and do not evaluate. (a) \(\int x \sin \left(x^{2}\right) d x\) (b) \(\int x^{2} \sin x d x\) (c) \(\int \frac{x^{2}}{1+x^{2}} d x\) (d) \(\int \frac{x}{\left(1+x^{2}\right)^{2}} d x\) (e) \(\int x^{3} e^{x^{2}} d x\) (f) \(\int \frac{\sin x}{2+\cos x} d x\)

Short Answer

Expert verified

(a) Substitution; (b) Not suitable; (c) Substitution; (d) Substitution; (e) Not suitable; (f) Substitution.

Step by step solution

Evaluate Integral (a)

For the integral \( \int x \sin(x^2) \, dx \), we perform substitution. Let \( u = x^2 \), then \( du = 2x \, dx \) or \( \frac{1}{2} du = x \, dx \). The integral becomes \( \frac{1}{2} \int \sin(u) \, du \), which evaluates to \( -\frac{1}{2} \cos(u) + C \). Substituting back \( u = x^2 \), the solution is \( -\frac{1}{2} \cos(x^2) + C \).

Evaluate Integral (b)

For \( \int x^2 \sin x \, dx \), substitution is not straightforward because the variable inside the sine function is not the same as the polynomial part. Integration by parts or another method is more suitable here. Therefore, substitution is not appropriate for evaluating this integral.

Evaluate Integral (c)

Consider \( \int \frac{x^2}{1+x^2} \, dx \). Here, substitution is suitable. Let \( u = 1 + x^2 \), then \( du = 2x \, dx \), but we need \( x^2 \, dx \), which can be rewritten in terms of \( u \) as: \( x^2 = u - 1 \). Thus, \( dx = \frac{1}{2x} du \) and the integral becomes \( \int \left(1 - \frac{1}{u}\right) du \). Upon integration, it yields \( u - \ln|u| + C \), which back-substituting gives \( 1+x^2 - \ln|1+x^2| + C \).

Evaluate Integral (d)

In \( \int \frac{x}{\left(1+x^2\right)^2} \, dx \), utilize substitution. Take \( u = 1 + x^2 \), hence \( du = 2x \, dx \), or \( \frac{1}{2} du = x \, dx \). Substitute to get \( \frac{1}{2} \int u^{-2} \, du \), which equates to \( -\frac{1}{2u} + C \). Plugging back, \( -\frac{1}{2(1+x^2)} + C \).

Evaluate Integral (e)

For \( \int x^3 e^{x^2} \, dx \), substitution is appropriate. Set \( u = x^2 \), so \( du = 2x \, dx \) and \( x^3 \, dx = x \, x^2 \, dx = \frac{1}{2} x u \, du \), which isn't neatly integrable by substitution. This integral is better solved by integration by parts. Thus, substitution is not appropriate.

Evaluate Integral (f)

For \( \int \frac{\sin x}{2 + \cos x} \, dx \), try substitution. Let \( u = 2 + \cos x \), then \( du = -\sin x \, dx \) implying \( -du = \sin x \, dx \). The integral becomes \( -\int \frac{1}{u} \, du \), which evaluates to \( -\ln|u| + C \). Substituting back gives \( -\ln|2 + \cos x| + C \).

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Key Concepts

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

Integration by Substitution

Integration by substitution is a powerful technique used in calculus to simplify the process of finding antiderivatives. It works by changing variables to make the integral simpler to solve. This technique is particularly useful when you have a complex function composed of another function and its derivative.To use integration by substitution, follow these steps:

Identify a part of the integrand that can be substituted with a single variable. This is usually the inner function.
Express the chosen part in terms of a new variable, \( u \), such that \( u = g(x) \) and thus, \( du = g'(x) \, dx \).
Rewrite the entire integral in terms of \( u \) and \( du \), which often simplifies the integral.
Perform the integration with respect to \( u \).
Finally, substitute back the original variable to get the solution in terms of \( x \).

This method was used effectively in the evaluation of integral (a):

Set \( u = x^2 \) leading to \( du = 2x \, dx \).
The integral \( \int x \sin(x^2) \, dx \) is transformed to \( \frac{1}{2} \int \sin(u) \, du \).
After integration, we get a simple antiderivative: \( -\frac{1}{2} \cos(u) + C \).
Substitute back \( u = x^2 \) to obtain the final result: \( -\frac{1}{2} \cos(x^2) + C \).

Integrals Involving Trigonometric Functions

Trigonometric functions are common in calculus, and integrating them often requires specific techniques or substitution methods. Recognizing patterns and identities can simplify the process significantly.For example, some integrals involve functions like sine and cosine, where substitutions simplify their integration. Let's consider example (f) from the exercise:

The integral \( \int \frac{\sin x}{2 + \cos x} \, dx \) had a trigonometric function in the denominator, which suggested trying substitution.
Let \( u = 2 + \cos x \). Then \( du = -\sin x \, dx \), which means \( -du = \sin x \, dx \).
This substitution simplifies the integral to \( -\int \frac{1}{u} \, du \).
The antiderivative is \( -\ln|u| + C \), and substituting back for \( u \) gives \( -\ln|2 + \cos x| + C \).

Identifying these patterns and using the right identity or substitution can greatly simplify the integration process.

Integration Techniques

In calculus, various techniques can be employed to solve integrals effectively. Depending on the complexity and nature of the function involved, choosing the right technique is essential.Here are a few common techniques:

Integration by Parts: Used when the integrand is a product of functions and substitution does not apply directly. Useful in integral (b), where \( \int x^2 \sin x \, dx \) might be handled better by this method.
Partial Fraction Decomposition: Useful for rational functions where the degree of the numerator is less than the denominator, making simple integrals with fractions easier.
Trigonometric Substitution: Effective for integrals involving roots of quadratic expressions alongside trigonometric identities.
Numerical Integration: In cases where finding an antiderivative analytically is complex, numerical methods allow approximation of integrals.

These techniques not only diversify your approach to solve integrals but also increase efficiency in evaluating complex integrals. Understanding when and how to apply each technique is crucial for mastering integration in calculus.

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Short Answer

Step by step solution

Evaluate Integral (a)

Evaluate Integral (b)

Evaluate Integral (c)

Evaluate Integral (d)

Evaluate Integral (e)

Evaluate Integral (f)

Key Concepts

Integration by Substitution

Integrals Involving Trigonometric Functions

Integration Techniques

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