Problem 26
Question
use the method of substitution to find each of the following indefinite integrals. $$ \int x^{2} \cos \left(x^{3}+5\right) d x $$
Step-by-Step Solution
Verified Answer
\(\frac{1}{3} \sin(x^3 + 5) + C\)
1Step 1: Identify the Substitution
Look for an inner function whose derivative is present in the integral. In this case, observe that the expression inside the cosine function is \( u = x^3 + 5 \).The derivative of this function is:\( \frac{du}{dx} = 3x^2 \).Notice how this is closely related to the term \( x^2 \) in the integral which suggests we can rewrite the integral in terms of \( u \).
2Step 2: Perform the Substitution
Given \( u = x^3 + 5 \), calculate \( du \) and solve for \( dx \):\[ du = 3x^2 \, dx \]Thus,\[ dx = \frac{du}{3x^2}. \]Now substitute \( x^3 + 5 \) with \( u \) and \( dx \) with \( \frac{du}{3x^2} \) in the integral:\[ \int x^2 \cos(u) \cdot \frac{du}{3x^2} \].
3Step 3: Simplify the Integral
Cancel out \( x^2 \) in the integral:\[ \int \frac{1}{3} \cos(u) \, du. \]This simplifies to:\[ \frac{1}{3} \int \cos(u) \, du. \]
4Step 4: Integrate with Respect to u
Find the indefinite integral of \( \cos(u) \) with respect to \( u \):\[ \int \cos(u) \, du = \sin(u) + C. \]So,\[ \frac{1}{3} \int \cos(u) \, du = \frac{1}{3} \sin(u) + C. \]
5Step 5: Back Substitute the Original Variable
Substitute back \( u = x^3 + 5 \) to express the integral in terms of \( x \):\[ \frac{1}{3} \sin(x^3 + 5) + C. \]
6Step 6: Final Result
The solution to the indefinite integral is:\[ \int x^{2} \cos(x^{3} + 5) \, dx = \frac{1}{3} \sin(x^3 + 5) + C. \]
Key Concepts
Indefinite IntegralsTrigonometric FunctionsIntegration Techniques
Indefinite Integrals
Indefinite integrals are a fundamental concept in calculus. They represent the collection of all antiderivatives of a function. Essentially, when you integrate a function indefinitely, there is no specified range or limits to restrict the computation, leading to a result that includes an arbitrary constant, typically denoted as \( C \).
Unlike definite integrals which yield a number, indefinite integrals yield a family of functions. These functions will differ by a constant. The notation for an indefinite integral is typically given as:
Unlike definite integrals which yield a number, indefinite integrals yield a family of functions. These functions will differ by a constant. The notation for an indefinite integral is typically given as:
- \( \int f(x) \, dx \)
Trigonometric Functions
Trigonometric functions are a set of functions that relate the angles and sides of triangles, predominant in various fields of mathematics, including calculus.
In the context of integration, trigonometric functions such as \( \sin(x) \), \( \cos(x) \), \( \tan(x) \), and their respective inverses and reciprocals are integrated to find their antiderivatives. The integral of \( \cos(x) \), for instance, leads us to its antiderivative, \( \sin(x) \), given by:
Handling integrals involving trigonometric functions often employs various integration techniques, including substitution. Recognizing trigonometric identities and their derivatives is essential for both evaluating integrals and understanding the patterns in integration results.
Understanding these functions and their properties aids not just in solving integrals but also in comprehending how functions interact with each other in calculus.
In the context of integration, trigonometric functions such as \( \sin(x) \), \( \cos(x) \), \( \tan(x) \), and their respective inverses and reciprocals are integrated to find their antiderivatives. The integral of \( \cos(x) \), for instance, leads us to its antiderivative, \( \sin(x) \), given by:
- \( \int \cos(u) \, du = \sin(u) + C \)
Handling integrals involving trigonometric functions often employs various integration techniques, including substitution. Recognizing trigonometric identities and their derivatives is essential for both evaluating integrals and understanding the patterns in integration results.
Understanding these functions and their properties aids not just in solving integrals but also in comprehending how functions interact with each other in calculus.
Integration Techniques
Integration techniques refer to various methods used to evaluate integrals, especially when finding an antiderivative is not straightforward. The method of substitution is a powerful technique often employed to simplify integrals.
Substitution involves replacing a specific expression in the function with a single variable to make integration more manageable. In our original problem, choosing \( u = x^3 + 5 \) is a strategic decision that simplifies the integral. By identifying derivatives present, we can rewrite the integral in terms of \( u \) and \( du \), leading to an easier form to integrate:
Once the integral is simplified and calculated, the process of back-substitution returns the substituted parts back to the original variable. This step ensures the final result is expressed in terms of the initial variables given in the problem.
Utilizing substitution effectively can greatly simplify complex integrals and is a valuable tool in the arsenal of calculus.
Substitution involves replacing a specific expression in the function with a single variable to make integration more manageable. In our original problem, choosing \( u = x^3 + 5 \) is a strategic decision that simplifies the integral. By identifying derivatives present, we can rewrite the integral in terms of \( u \) and \( du \), leading to an easier form to integrate:
- \( \int x^2 \cos(u) \cdot \frac{du}{3x^2} = \frac{1}{3} \int \cos(u) \, du \)
Once the integral is simplified and calculated, the process of back-substitution returns the substituted parts back to the original variable. This step ensures the final result is expressed in terms of the initial variables given in the problem.
Utilizing substitution effectively can greatly simplify complex integrals and is a valuable tool in the arsenal of calculus.
Other exercises in this chapter
Problem 26
Find \(G^{\prime}(x)\). $$ G(x)=\int_{\cos x}^{\sin x} t^{5} d t $$
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Find all values of \(c\) that satisfy the Mean Value Theorem for Integrals on the given interval. $$ S(y)=y^{2} ; \quad[0, b] $$
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The velocity function for an object is given. Assuming that the object is at the origin at time \(t=0\), find the position at time \(t=4\). $$ v(t)=\left\\{\beg
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Find the interval(s) on which the graph of \(y=f(x), x \geq 0\), is (a) increasing, and (b) concave up. $$ f(x)=\int_{0}^{x} \frac{s}{\sqrt{1+s^{2}}} d s $$
View solution