Understanding the Integration Problem

We are asked to evaluate the indefinite integral:

\(\int (e^{\log x} + \sin x) \cos x \, dx\)

The first step is to simplify the expression inside the integral. Recall the property of logarithms and exponentials that \(e^{\log x} = x\) for \(x > 0\). Using this property, the integral becomes:

\(\int (x + \sin x) \cos x \, dx\)

Now, we can distribute \(\cos x\) across the terms inside the parenthesis:

\(\int (x \cos x + \sin x \cos x) \, dx\)

This integral can be split into two separate integrals:

\(\int x \cos x \, dx + \int \sin x \cos x \, dx\)

Solving the First Integral: \( \int x \cos x \, dx \)

The first integral \(\int x \cos x \, dx\) requires the technique of integration by parts. The formula for integration by parts is:

\(\int u \, dv = uv - \int v \, du\)

We need to choose appropriate parts for \(u\) and \(dv\). A common guideline (LIATE/ILATE) suggests letting \(u\) be a term that simplifies when differentiated and \(dv\) be a term that is easily integrated.

• Let \(u = x\)
• Let \(dv = \cos x \, dx\)

Now, we find \(du\) and \(v\):

• Differentiating \(u = x\) gives \(du = 1 \, dx\) or just \(du = dx\).
• Integrating \(dv = \cos x \, dx\) gives \(v = \int \cos x \, dx = \sin x\).

Now, substitute these into the integration by parts formula:

\(\int x \cos x \, dx = (x)(\sin x) - \int (\sin x) \, dx\)

\(\int x \cos x \, dx = x \sin x - (-\cos x) + C_1\)

\(\int x \cos x \, dx = x \sin x + \cos x + C_1\)

Here, \(C_1\) is the constant of integration for the first integral.

Solving the Second Integral: \( \int \sin x \cos x \, dx \)

The second integral \(\int \sin x \cos x \, dx\) can be solved using a simple substitution method.

Let \(u = \sin x\).

Then, the differential \(du\) is the derivative of \(\sin x\) multiplied by \(dx\):

\(du = \cos x \, dx\)

Now, substitute \(u\) and \(du\) into the integral:

\(\int \sin x \cos x \, dx = \int u \, du\)

This is a basic power rule integral:

\(\int u \, du = \dfrac{u^2}{2} + C_2\)

Substitute back \(u = \sin x\):

\(\int \sin x \cos x \, dx = \dfrac{\sin^2 x}{2} + C_2\)

Here, \(C_2\) is the constant of integration for the second integral.

Combining the Results

The original integral is the sum of the two integrals we just solved:

\(\int (e^{\log x} + \sin x) \cos x \, dx = \int x \cos x \, dx + \int \sin x \cos x \, dx\)

Substitute the results from our calculations:

\(\int (e^{\log x} + \sin x) \cos x \, dx = (x \sin x + \cos x + C_1) + \left(\dfrac{\sin^2 x}{2} + C_2\right)\)

Combine the constants of integration \(C_1\) and \(C_2\) into a single constant \(C = C_1 + C_2\):

\(\int (e^{\log x} + \sin x) \cos x \, dx = x \sin x + \cos x + \dfrac{\sin^2 x}{2} + C\)

Comparing with Options

Let's compare our derived solution with the given options:

• Option 1: \(\rm \sin x + x \cos x + \dfrac{\sin^2 x}{2}+c\) (Incorrect - terms are similar but sign might be an issue, check carefully)
• Option 2: \(\rm \sin x - x \cos x + \dfrac{\sin^2 x}{2}+c\) (Incorrect)
• Option 3: \(\rm x \sin x + \cos x + \dfrac{\sin^2 x}{2}+c\) (Matches our result)
• Option 4: \(\rm x \sin x - x \cos x + \dfrac{\sin^2 x}{2}+c\) (Incorrect)

Our calculated result, \(x \sin x + \cos x + \dfrac{\sin^2 x}{2} + C\), matches Option 3.

Revision Table: Key Integration Techniques

Additional Information: Properties Used

This integration problem utilized a few key mathematical concepts:

• Logarithm and Exponential Property: The fundamental property \(e^{\log x} = x\) (for \(x > 0\)). This simplifies the initial expression significantly.
• Linearity of Integration: The property that \(\int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx\). This allowed us to split the integral into two more manageable parts.
• Basic Trigonometric Integrals: Knowing the standard integrals like \(\int \cos x \, dx = \sin x\) and \(\int \sin x \, dx = -\cos x\).
• Trigonometric Identities (Optional approach for second integral): Although we used substitution, the identity \(\sin(2x) = 2 \sin x \cos x\) could also have been used for \(\int \sin x \cos x \, dx\). \(\int \sin x \cos x \, dx = \int \frac{1}{2} \sin(2x) \, dx = -\frac{1}{4} \cos(2x) + C'\). This result is equivalent to \(\frac{\sin^2 x}{2} + C_2\) because \(\sin^2 x = \frac{1 - \cos(2x)}{2}\), so \(\frac{\sin^2 x}{2} = \frac{1 - \cos(2x)}{4} = \frac{1}{4} - \frac{\cos(2x)}{4}\). The difference is just a constant, which is absorbed into the integration constant.