Question

Compute the definite integrals. Use a graphing utility to confirm your answers. Evaluate \(\int \cos x \ln (\sin x) d x\)

Short answer

The integral evaluates to \( \ln(\sin x) \cdot \sin x - \sin x + C \).

Step-by-step solution

Identify the Integral

The integral we have is \( \int \cos x \ln (\sin x) \, dx \). This is a product of two functions: \(\cos x\) and \(\ln(\sin x)\).

Integration by Parts Setup

For integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). Let's choose \( u = \ln(\sin x) \) and \( dv = \cos x \, dx \). This means \( du = \frac{1}{\sin x} \cdot \cos x \, dx \) and \( v = \sin x \).

Apply Integration by Parts

Substitute into the integration by parts formula: \( \int \ln(\sin x) \cdot \cos x \, dx = \ln(\sin x) \cdot \sin x - \int \sin x \cdot \frac{1}{\sin x} \cdot \cos x \, dx \).

Simplify the Remaining Integral

The remaining integral simplifies to \( \int \cos x \, dx \), because \( \sin x \cdot \frac{1}{\sin x} = 1 \).

Compute Final Integral

Compute the integral \( \int \cos x \, dx = \sin x + C \).

Combine Results

Combine the results from the integration by parts: \( \ln(\sin x) \cdot \sin x - \sin x + C \).

Graphing Utility Confirmation

Use a graphing utility to confirm the integral is \( \ln(\sin x) \cdot \sin x - \sin x + C \) by comparing the graph of the original function with its antiderivative. This ensures accuracy of the result.

Key concepts

Integration by Parts

Integration by parts is a mathematical technique useful when you're trying to integrate products of functions. This method is derived from the product rule of differentiation and is expressed by the formula:\[ \int u \, dv = uv - \int v \, du \]When you have two functions multiplied together inside an integral, like in \( \int \cos x \ln (\sin x) \, dx \), you can apply integration by parts. In this example:

• Choose \( u = \ln(\sin x) \)
• Then, \( dv = \cos x \, dx \)

The differentiation of \( u \) gives\( du = \frac{1}{\sin x} \cdot \cos x \, dx \), and by integrating \( dv \), we get \( v = \sin x \). Substituting them back into the formula helps simplify the integral, transforming it into a potentially solvable form. Just be sure to simplify the remaining integrals carefully.

Product of Functions

In calculus, the product of functions involves two or more functions being multiplied. These types of integrals can be tricky without the right techniques, such as integration by parts. Consider an expression like \( \cos x \ln(\sin x) \), which consists of the trigonometric function \( \cos x \) and the logarithmic function \( \ln(\sin x) \). When faced with such products in integration, you can't easily separate or rearrange them without altering the integral's meaning. Instead, applying integration by parts allows us to strategically break down and evaluate these products through a series of simpler integration steps. Choosing which function to differentiate and which to integrate is crucial for ease of computation. Often, you select the function that becomes simpler when differentiated as "\( u \)", and the rest as "\( dv \)".

Graphing Utility

Graphing utilities are powerful tools that visually verify the correctness of your integrals and their solutions. After analytically solving an integral, like\( \int \cos x \ln (\sin x) \, dx \), it's wise to use technology to back up your findings.Many calculators and software, such as Desmos or GeoGebra, offer the capability to graph both the original function and its antiderivative. By analyzing and comparing the graphs or checking numerical values at certain points, you can confirm that your solution aligns with what the graphing utility produces.

• Helps in verifying complex solutions
• Provides a visual understanding of the integral
• Assists in detecting any errors made during manual calculations

Antiderivative

An antiderivative of a function is simply another function whose derivative returns the original function. In the context of integrals, finding the antiderivative is equivalent to evaluating the integral itself. For definite integrals, you would find the antiderivative and then assess it over the given interval.In our problem, calculating \( \int \cos x \ln (\sin x) \, dx \) yields an antiderivative of the form \( \ln(\sin x) \cdot \sin x - \sin x + C \). The constant \( C \) represents any constant of integration, since derivatives of constants are zero and thus, they affect the antiderivative but not its derivative.By knowing the antiderivative, you can:

• Calculate exact areas under curves with definite integrals
• Understand the accumulation of quantities represented by your function
• Readily work with the functions in calculus and physics problems

The antiderivative also equips you with a tool to analytically solve and understand integration problems that arise across various disciplines.