Question

apply integration by parts twice to evaluate each integral. $$ \int \sin (\ln x) d x $$

Short answer

\(\int \sin(\ln x) \, dx = \frac{1}{2}(x \sin(\ln x) - x \cos(\ln x)) + C\).

Step-by-step solution

Choose u and dv

In integration by parts, the formula is \[\int u \ dv = uv - \int v \ du.\]Here, we choose: - Let \( u = \sin (\ln x) \) \( \Rightarrow du = \cos (\ln x) \frac{1}{x} dx \).- Let \( dv = dx = \frac{d}{dx} x \Rightarrow v = x \).

Apply Integration by Parts First Time

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

Choose u and dv for Second Integration

To evaluate \(\int \cos(\ln x) \, dx\) using integration by parts again:- Let \( u = \cos(\ln x) \) with \(du = -\sin(\ln x)\frac{1}{x}dx\).- Let \(dv = dx = \frac{d}{dx} x \Rightarrow v = x\).

Apply Integration by Parts Second Time

Substitute into the integration by parts formula again:\[\int \cos(\ln x) \, dx = x \cos(\ln x) - \int x (-\sin(\ln x)) \frac{1}{x} \, dx.\]Simplifying the integral:\[\int \cos(\ln x) \, dx = x \cos(\ln x) + \int \sin(\ln x) \, dx.\]

Solve the Equation for the Original Integral

Substitute the result from Step 4 into the equation from Step 2:\[\int \sin(\ln x) \, dx = x \sin(\ln x) - \left(x \cos(\ln x) + \int \sin(\ln x) \, dx\right).\]Rearrange to solve for the original integral:\[2 \int \sin(\ln x) \, dx = x \sin(\ln x) - x \cos(\ln x).\]

Divide Both Sides to Solve for the Integral

Finally, divide both sides by 2:\[\int \sin(\ln x) \, dx = \frac{1}{2} \left( x \sin(\ln x) - x \cos(\ln x) \right) + C,\]where \(C\) is the constant of integration.

Key concepts

Trigonometric Integrals

Trigonometric integrals involve integrating functions that contain trigonometric expressions. These integrals often encompass angles or logarithmic transformations, like \( \sin(\ln x) \) and \ \cos(\ln x) \, which are the primary focus in this context. Integrating these types of functions can be challenging, but they become manageable through the use of different calculus techniques, such as integration by parts.
When dealing with trigonometric integrals:

• It's essential to remember that the goal is to simplify the integral into a more easily solvable form.
• Applying techniques like substitution, trigonometric identities, and in this case, integration by parts can help in solving them.
• Trigonometric integrals often involve re-writing the function in a way that makes it easier to integrate.

In our exercise, we applied integration by parts twice. This approach is particularly useful when direct integration is not straightforward.

Integration Techniques

Integration techniques are strategies to evaluate integrals that are otherwise challenging to solve using standard integration methods. Knowing when and how to apply these techniques is crucial. In this exercise, integration by parts was the chosen technique. This method is particularly useful when dealing with products of functions, which can complicate direct integration.
Integration by Parts follows the formula:
\[ \int u \, dv = uv - \int v \, du \]

• This method transforms the original integral into a product of functions minus another integral.
• To effectively use integration by parts, it requires selecting two parts from the function being integrated: \( u \, and \, dv \).
• The challenge is to choose \( u \) and \, dv \, such that the resulting \( \int v \, du \) is simpler to solve than the original integral.

In the given exercise, integration by parts was applied twice to manage the trigonometric functions and simplify the calculation.

Calculus

Calculus is the branch of mathematics that deals with the properties of derivatives and integrals of functions. It provides the tools necessary to analyze rates of change and find areas under curves, among other applications. The fundamental theorem of calculus connects differentiation with integration, laying the foundation for solving complex mathematical problems.

• Understanding calculus principles is crucial as they evolve from simple derivative calculations to complex integrals.
• Techniques like integration by parts, substitution, and trigonometric identities are essential for tackling advanced calculus problems.
• Thanks to these tools, we can handle functions involving both algebraic and trigonometric expressions, as seen in the exercise solving \( \int \sin(\ln x) \, dx \).

Mastering these concepts is key for students aiming to excel in calculus and for real-world applications in physics, engineering, and technology.