Problem 92
Question
Assertion: \(\int e^{(x \sin x+\cos x)}\left(\frac{x^{2} \cos ^{2} x-x \sin x-\cos x}{x^{2}}\right) d x\) \(=e^{(x \sin x+\cos x)} \cdot \frac{\cos x}{x}+C\) Reason: \(\int e^{g(x)}\left\\{f(x) g^{\prime}(x)+f^{\prime}(x)\right\\} d x\) \(=e g^{\prime} x^{\prime} f(x)\)
Step-by-Step Solution
Verified Answer
The assertion is correct; the integration matches using the given reason formula.
1Step 1: Identify the Parts of the Assertion
The assertion states that the integral \(\int e^{(x \sin x+\cos x)}\left(\frac{x^{2} \cos ^{2} x-x \sin x-\cos x}{x^{2}}\right) d x\) is equal to \(e^{(x \sin x+\cos x)} \cdot \frac{\cos x}{x}+C\). We need to identify the function \(f(x)\) and its derivative \(f'(x)\) from this expression.
2Step 2: Align with the Reason Formula
The given reason formula is \(\int e^{g(x)}\{f(x) g^{\prime}(x)+f^{\prime}(x)\} dx = e^{g(x)} f(x)\). We need \(g(x) = x \sin x + \cos x\), and \(g'(x)\) needs to be calculated.
3Step 3: Differentiate to Find \(g'(x)\)
Calculate the derivative \(g'(x)\) of \(g(x) = x \sin x + \cos x\). Use the product rule: \(g'(x) = (x \sin x)' + (\cos x)' = x \cos x + \sin x - \sin x = x \cos x\).
4Step 4: Find \(f(x)\) and \(f'(x)\)
Looking at the integral: \[e^{g(x)}\left(\frac{x^{2} \cos^{2} x - x \sin x - \cos x}{x^{2}}\right)\], let \(f(x) = \frac{\cos x}{x}\). Then \(f'(x) = -\frac{\sin x \cdot x - \cos x}{x^2}\).
5Step 5: Verify the Formula Matching
Check if the given integral matches with the reason formula: Substitute \(f(x)\) and \(g'(x)\) into \(f(x) g'(x) + f'(x)\). This results in \(\frac{\cos x}{x} \cdot x \cos x + \left(-\frac{x \sin x - \cos x}{x^2}\right)\), which simplifies to \(\frac{x^{2} \cos^{2} x - x \sin x - \cos x}{x^2}\), as in the integral.
6Step 6: Confirm Assertion
Since we've verified that applying the given formula results in the integral matching the assertion, the assertion is indeed correct. The integration simplifies correctly to \(e^{g(x)} f(x) + C = e^{(x \sin x + \cos x)} \cdot \frac{\cos x}{x} + C\).
Key Concepts
trigonometric integrationdifferentiationintegration technique
trigonometric integration
Trigonometric integration is a technique used in calculus when integrating functions involving trigonometric terms. In this context, we're dealing with functions like \(\cos x\) and \(\sin x\) within the integral. Understanding how these trigonometric functions interact within integrals often requires knowledge of both trigonometric identities and integration rules.
For instance, in the provided solution, we see \(x \sin x + \cos x\) appearing within the exponent of an exponential function. This is a part of what constitutes the function \(g(x)\), which is crucial for the integration process. The challenge in trigonometric integration involves identifying patterns or familiar forms to strategically apply relevant techniques to simplify and evaluate the integral.
In cases such as this, break down the integral into manageable parts and utilize known calculus strategies, like integration by parts or substitution, to solve the problem efficiently. Recognize the interactions between terms to spot trigonometric identities that simplify expressions within the integral.
For instance, in the provided solution, we see \(x \sin x + \cos x\) appearing within the exponent of an exponential function. This is a part of what constitutes the function \(g(x)\), which is crucial for the integration process. The challenge in trigonometric integration involves identifying patterns or familiar forms to strategically apply relevant techniques to simplify and evaluate the integral.
In cases such as this, break down the integral into manageable parts and utilize known calculus strategies, like integration by parts or substitution, to solve the problem efficiently. Recognize the interactions between terms to spot trigonometric identities that simplify expressions within the integral.
differentiation
Differentiation is a fundamental concept in calculus that refers to finding the derivative of a function. The derivative represents the rate of change of a function with respect to a variable. In the provided exercise, differentiation plays a central role because we need to determine the derivative \(g'(x)\) of the function \(g(x) = x \sin x + \cos x\).
To find \(g'(x)\), we apply the product rule, which states that the derivative of a product of two functions is given by \((u \cdot v)' = u' \cdot v + u \cdot v'\). Here, for the function \(x \sin x\), let \(u = x\) and \(v = \sin x\), so \(u' = 1\) and \(v' = \cos x\). Then, \((x \sin x)' = x \cdot \cos x + \sin x\). Adding \(\cos x\)'s derivative, which is \(-\sin x\), we simplify to \(g'(x) = x \cos x\).
Understanding differentiation is essential as it allows us to identify \(g'(x)\), a key part of our integration problem, ensuring our steps align with the integration formula.
To find \(g'(x)\), we apply the product rule, which states that the derivative of a product of two functions is given by \((u \cdot v)' = u' \cdot v + u \cdot v'\). Here, for the function \(x \sin x\), let \(u = x\) and \(v = \sin x\), so \(u' = 1\) and \(v' = \cos x\). Then, \((x \sin x)' = x \cdot \cos x + \sin x\). Adding \(\cos x\)'s derivative, which is \(-\sin x\), we simplify to \(g'(x) = x \cos x\).
Understanding differentiation is essential as it allows us to identify \(g'(x)\), a key part of our integration problem, ensuring our steps align with the integration formula.
integration technique
Integration techniques are methods used to solve integrals more effectively. In this exercise, integration by parts emerges as a crucial technique. It's used to integrate products of functions by transforming the integral into a simpler one. This method is based on the rule \(\int u \cdot dv = u \cdot v - \int v \cdot du\), which is derived from the product rule for differentiation.
However, the task at hand utilizes a specific formula \(\int e^{g(x)} \{f(x) g'(x) + f'(x)\} dx = e^{g(x)} f(x) + C\), which simplifies the problem due to its unique structure. Identifying suitable \(f(x)\) and \(f'(x)\) is critical: here, \(f(x) = \frac{\cos x}{x}\) is selected, and its derivative, \(f'(x) = -\frac{x \sin x - \cos x}{x^2}\), aligns with our problem's expressions.
The key to mastering integration techniques is practice and familiarity with different types: substitution, partial fractions, trigonometric identities, and the use of specific formulas like the one here. Each technique provides a tactical advantage, allowing you to tackle otherwise complex integrals in simpler steps.
However, the task at hand utilizes a specific formula \(\int e^{g(x)} \{f(x) g'(x) + f'(x)\} dx = e^{g(x)} f(x) + C\), which simplifies the problem due to its unique structure. Identifying suitable \(f(x)\) and \(f'(x)\) is critical: here, \(f(x) = \frac{\cos x}{x}\) is selected, and its derivative, \(f'(x) = -\frac{x \sin x - \cos x}{x^2}\), aligns with our problem's expressions.
The key to mastering integration techniques is practice and familiarity with different types: substitution, partial fractions, trigonometric identities, and the use of specific formulas like the one here. Each technique provides a tactical advantage, allowing you to tackle otherwise complex integrals in simpler steps.
Other exercises in this chapter
Problem 89
Column-I Column-II (A) \(\int \sqrt{\sec x-1} d x\) 1\. \(\sin ^{-1}(\tan x)\) (B) \(\int \frac{d x}{\cos x \sqrt{\cos 2 x}}\) 2\. \(\sec ^{-1}(\sec x+\cos x)\)
View solution Problem 90
Column-I Column-II (A) \(\int \frac{d x}{(1+\sqrt{x}) \sqrt{x-x^{2}}}\) 1\. \(\frac{2(\sqrt{x}-1)}{\sqrt{1-x}}\) (B) \(\int\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\r
View solution Problem 95
Assertion: \(\begin{aligned} \int x^{x+1} \log x(1+\log x) d x \\ &=x^{x}(x \log x-1)+C \end{aligned}\) Reason: \(\int x^{x}(1+\log x) d x=x^{x}\)
View solution Problem 96
\(\int \frac{d x}{x\left(x^{n}+1\right)}\) is equal to: (A) \(\frac{1}{n} \log \left(\frac{x^{\prime \prime}}{x^{n}+1}\right)+c\) (B) \(\frac{1}{n} \log \left(\
View solution