Problem 68
Question
Find an antiderivative by reversing the chain rule, product rule or quotient rule. $$\int\left(\sqrt{x^{2}+1} \cos x+\frac{x}{\sqrt{x^{2}+1}} \sin x\right) d x$$
Step-by-Step Solution
Verified Answer
The antiderivative is \(\arcsin\left(\frac{x}{\sqrt{x^{2}+1}}\right) + C\) where C is the integration constant.
1Step 1: Identify the two superimposed integrals
We can rewrite the integral as the sum of two integrals:\[ \int (\sqrt{x^{2}+1} \cos x) dx + \int \frac{x}{\sqrt{x^{2}+1}} \sin x dx.\]
2Step 2: Recognize application of the Chain rule
The chain rule states that \(d[f(g(x))]=f'(g(x))g'(x)dx\). If we consider \(f(u) = \sin(u)\) and \(g(x) = \arcsin\left(\frac{x}{\sqrt{x^{2}+1}}\right)\), we can notice that the second integral follows from applying the chain rule.
3Step 3: Recognize the derivative
From the previous step, realize that the whole integrand is the derivative of \(\arcsin\left(\frac{x}{\sqrt{x^{2}+1}}\right)\). Therefore, you can directly conclude that the antiderivative is \(\arcsin\left(\frac{x}{\sqrt{x^{2}+1}}\right)\) plus constant C.
Key Concepts
Chain Rule in CalculusProduct Rule in DifferentiationQuotient Rule for Handling DivisionsIntegral Calculus and Antiderivatives
Chain Rule in Calculus
The chain rule is a fundamental principle in calculus used when taking the derivative of a composite function. It involves taking the derivative of the outer function and multiplying it by the derivative of the inner function. In layman's terms, if one function is nested inside another, the chain rule enables us to differentiate the entire expression in a step-by-step fashion.
For example, if we have a function defined as \( h(x) = f(g(x)) \), the chain rule dictates that the derivative of \( h(x) \) with respect to \( x \) is \( h'(x) = f'(g(x)) \cdot g'(x) \). In the context of antiderivatives, we reverse this process, looking for a function whose derivative appears in the integrand, and identify matching patterns that suggest the use of the chain rule in reverse.
For example, if we have a function defined as \( h(x) = f(g(x)) \), the chain rule dictates that the derivative of \( h(x) \) with respect to \( x \) is \( h'(x) = f'(g(x)) \cdot g'(x) \). In the context of antiderivatives, we reverse this process, looking for a function whose derivative appears in the integrand, and identify matching patterns that suggest the use of the chain rule in reverse.
Product Rule in Differentiation
The product rule comes into play when we want to differentiate an expression that is the product of two or more functions. The rule states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
In notation, if \( u(x) \) and \( v(x) \) are two differentiable functions, then the derivative of their product is given by \( (u \cdot v)' = u' \cdot v + u \cdot v' \). This is an essential concept when solving integrals that involve products of functions, as sometimes it might be clearer to see how two functions were multiplied together before differentiating.
In notation, if \( u(x) \) and \( v(x) \) are two differentiable functions, then the derivative of their product is given by \( (u \cdot v)' = u' \cdot v + u \cdot v' \). This is an essential concept when solving integrals that involve products of functions, as sometimes it might be clearer to see how two functions were multiplied together before differentiating.
Quotient Rule for Handling Divisions
The quotient rule is used when differentiating a function that is the quotient of two functions. It is slightly more complex than the product rule and states that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
In symbolic form, for two functions \( u(x) \) and \( v(x) \) where \( v(x) \) is not zero, the quotient rule is expressed as \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). Recognizing when the quotient rule has been applied is crucial for determining antiderivatives of rational functions, particularly during the process of integration.
In symbolic form, for two functions \( u(x) \) and \( v(x) \) where \( v(x) \) is not zero, the quotient rule is expressed as \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). Recognizing when the quotient rule has been applied is crucial for determining antiderivatives of rational functions, particularly during the process of integration.
Integral Calculus and Antiderivatives
Integral calculus is the branch of mathematics focused on finding the antiderivatives, or integrals, of functions. An antiderivative of a function \( f(x) \) is another function \( F(x) \) whose derivative is \( f(x) \). The process of finding antiderivatives is known as integration.
When integrating, we often look for a function whose derivative would give us the integrand. Reverse application of differentiation rules such as the chain rule, product rule, and quotient rule can be instrumental in this process. For instance, finding the antiderivative in the given exercise involved recognizing elements of the chain rule in the integrand. By associating each term in the given integral with its differentiation counterpart, we can effectively 'undo' the differentiation and find the original function, often including an arbitrary constant \( C \) since differentiation omits constants.
When integrating, we often look for a function whose derivative would give us the integrand. Reverse application of differentiation rules such as the chain rule, product rule, and quotient rule can be instrumental in this process. For instance, finding the antiderivative in the given exercise involved recognizing elements of the chain rule in the integrand. By associating each term in the given integral with its differentiation counterpart, we can effectively 'undo' the differentiation and find the original function, often including an arbitrary constant \( C \) since differentiation omits constants.
Other exercises in this chapter
Problem 68
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