Chapter 2: Problem 1
Integrating \(\int \cos ^{\prime} x \sin x d x\) Evaluate \(\int \cos ^{3} x \sin x d x\)
Short Answer
Expert verified
-\frac{\cos^4 x}{4} + C
Step by step solution
01
Recognize the Integral Form
The integral \( \int \cos^3 x \sin x \, dx \) is in a form where we need to make a substitution to simplify the integration process. Notice that the expression \( \cos^3 x \) can be tackled by expressing it as a function of some variable, while \( \sin x \) is the derivative of \( \cos x \).
02
Substitution Technique
Use the substitution \( u = \cos x \). This implies that the derivative \( \frac{du}{dx} = -\sin x \) or equivalently, \( du = -\sin x \, dx \). So, we can rewrite \( \sin x \, dx = -du \).
03
Substitute in the Integral
Applying the substitution, the integral becomes: \( \int \cos^3 x \sin x \, dx = \int u^3 (-du) = -\int u^3 \, du\).
04
Integrate the Simplified Form
Now integrate \( -\int u^3 \, du \). The antiderivative of \( u^3 \) is \( \frac{u^4}{4} \), so the integral becomes: \( -\frac{u^4}{4} + C \), where \( C \) is the constant of integration.
05
Substitute Back to Original Variable
Since we used \( u = \cos x \) in our substitution, revert back to the original variable by replacing \( u \). This gives us: \( -\frac{(\cos x)^4}{4} + C\). Therefore, the evaluated integral is: \( -\frac{\cos^4 x}{4} + C\).
06
Final Answer
The evaluated integral, when put back in terms of the original variable, is \( -\frac{\cos^4 x}{4} + C\).
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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 simplifying complex integrals. It's especially useful when an integral contains a function and its derivative. This method involves changing variables to make the integral easier to solve.
In the example given, we were dealing with the integral \( \int \cos^3 x \sin x \, dx \). Here, you recognize that the integral involves \( \cos x \) and its derivative \( \sin x \). Therefore, the substitution \( u = \cos x \) is a smart choice. This transforms the differential to \( du = -\sin x \, dx \), allowing us to simplify the original integral into \( -\int u^3 \, du \).
This simplification often leads the way to finding the solution to otherwise complicated problems. By choosing the right substitution, mathematicians can turn what seems like an insurmountable task into something much more manageable.
In the example given, we were dealing with the integral \( \int \cos^3 x \sin x \, dx \). Here, you recognize that the integral involves \( \cos x \) and its derivative \( \sin x \). Therefore, the substitution \( u = \cos x \) is a smart choice. This transforms the differential to \( du = -\sin x \, dx \), allowing us to simplify the original integral into \( -\int u^3 \, du \).
This simplification often leads the way to finding the solution to otherwise complicated problems. By choosing the right substitution, mathematicians can turn what seems like an insurmountable task into something much more manageable.
Trigonometric Integrals
Trigonometric integrals involve integrating expressions containing trigonometric functions like sine, cosine, and tangent. These types of integrals can be quite tricky because they often involve expressions raised to a power or a combination of multiple trig functions.
In our case, the integral \( \int \cos^3 x \sin x \, dx \) is a typical example. To solve it, we use trigonometric identities and the substitution method. Here, understanding the relationship between \( \sin x \) and \( \cos x \) is key. Since the integral involves powers of cosine, expressing it in a simpler form with respect to \( u \) as \( \int u^3 \, du \) makes the solution straightforward.
Trigonometric integrals require a good grasp of trigonometric identities and derivatives. Knowing how these functions transform and interact with one another allows you to simplify the integration process effectively.
In our case, the integral \( \int \cos^3 x \sin x \, dx \) is a typical example. To solve it, we use trigonometric identities and the substitution method. Here, understanding the relationship between \( \sin x \) and \( \cos x \) is key. Since the integral involves powers of cosine, expressing it in a simpler form with respect to \( u \) as \( \int u^3 \, du \) makes the solution straightforward.
Trigonometric integrals require a good grasp of trigonometric identities and derivatives. Knowing how these functions transform and interact with one another allows you to simplify the integration process effectively.
Antiderivatives
Antiderivatives, also known as indefinite integrals, represent the reverse process of differentiation. When you find an antiderivative, you are looking for a function whose derivative is the given function.
In the exercise, once the integral was transformed into \( -\int u^3 \, du \) through substitution, calculating the antiderivative of \( u^3 \) was straightforward. The antiderivative is \( \frac{u^4}{4} \), representing the function whose derivative would return the original \( u^3 \).
After integrating, always remember to include the constant of integration \( C \). This constant represents the family of all possible antiderivatives, reflecting the fact that derivatives of constants are zero.
In the exercise, once the integral was transformed into \( -\int u^3 \, du \) through substitution, calculating the antiderivative of \( u^3 \) was straightforward. The antiderivative is \( \frac{u^4}{4} \), representing the function whose derivative would return the original \( u^3 \).
After integrating, always remember to include the constant of integration \( C \). This constant represents the family of all possible antiderivatives, reflecting the fact that derivatives of constants are zero.
Calculus Problems
Calculus problems often involve finding the integral or derivative of functions, which requires a solid understanding of fundamental calculus principles. These problems can vary widely in complexity, from basic derivative calculations to intricate integration challenges.
In this particular integration example, several calculus techniques are employed, such as the substitution method and understanding the role of trigonometric functions and their derivatives. These skills are crucial for solving calculus problems and rely heavily on familiarity with integral calculus concepts.
For students tackling calculus problems, practice is essential. The more you practice different types of problems, the more adept you become at selecting and applying the appropriate methods. It enhances mathematical maturity and problem-solving skills, which are essential for success in advanced mathematical topics.
In this particular integration example, several calculus techniques are employed, such as the substitution method and understanding the role of trigonometric functions and their derivatives. These skills are crucial for solving calculus problems and rely heavily on familiarity with integral calculus concepts.
For students tackling calculus problems, practice is essential. The more you practice different types of problems, the more adept you become at selecting and applying the appropriate methods. It enhances mathematical maturity and problem-solving skills, which are essential for success in advanced mathematical topics.
