Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int \arctan (1 / t) d t $$

Short answer

\( \int \arctan(1/t) \, dt = t \cdot \arctan(1/t) + \frac{1}{2} \ln(t^2 + 1) + C \)

Step-by-step solution

Identify the Parts of Integration by Parts Formula

According to the integration by parts formula: \[\int u \, dv = uv - \int v \, du\]We need to identify which part of the integral will be \(u\) and which part will be \(dv\). Let's set \(u = \arctan (1/t)\) and \(dv = dt\).

Differentiate and Integrate the Parts

Differentiate \(u\) with respect to \(t\): \[ du = \frac{-1}{1 + (1/t)^2} \cdot \frac{1}{t^2} \, dt = - \frac{1}{t^2 + 1} \, dt \]Integrate \(dv\) to get \(v\):\[ v = t \]

Apply the Integration by Parts Formula

Substitute \(u\), \(du\), \(v\), and \(dv\) into the integration by parts formula:\[\int \arctan(1/t) \, d t = t \cdot \arctan(1/t) - \int t \left( - \frac{1}{t^2 + 1} \right) \, d t\]This simplifies to:\[\int \arctan(1/t) \, d t = t \cdot \arctan(1/t) + \int \frac{t}{t^2 + 1} \, d t\]

Solve the Remaining Integral

Now handle the remaining integral: \[\int \frac{t}{t^2 + 1} \, d t\]Let \( w = t^2 + 1 \), then \( dw = 2t \, dt \), or \( \, dt = \frac{1}{2t} \, dw \).The integral becomes:\[\int \frac{t}{w} \cdot \frac{1}{2t} \, dw = \frac{1}{2} \int \frac{1}{w} \, dw = \frac{1}{2} \ln |w| + C = \frac{1}{2} \ln (t^2 + 1) + C\]

Combine Results to Formulate the Final Answer

Substitute back to get the final answer: \[\int \arctan(1/t) \, d t = t \cdot \arctan(1/t) + \frac{1}{2} \ln(t^2 + 1) + C\]

Key concepts

Calculus

Calculus is a branch of mathematics that studies continuous change. It breaks down into two major fields: differential calculus and integral calculus. While differential calculus focuses on the concept of a derivative, integral calculus centers on the integral.
The goal of calculus is to help us understand how quantities change over time. For example, the rate at which a car's speed changes can be analyzed using differential calculus, whereas the total distance traveled over time can be deduced with integral calculus.
In the context of this exercise, calculus is applied to solve an integral using the method of integration by parts. This approach reflects the power of calculus in breaking complex problems into manageable parts.
Understanding calculus is crucial for fields like physics and engineering, where it assists in modeling dynamic systems.

Definite Integral

A definite integral calculates the accumulated value over an interval. Unlike indefinite integrals, which include a constant of integration, definite integrals result in a number interpretable as the area under a curve between two bounds.
This exercise on integration by parts involves an indefinite integral, yet understanding definite integrals provides context for the technique. Knowing how to convert between these two forms is essential for solving real-world problems.

• Limits of Integration: The bounds that define the interval over which you're integrating.
• Area Interpretation: Visualize definite integrals as measuring the area under a curve over a specific interval.
• Fundamental Theorem of Calculus: Links differentiation with integration, showing that they are inverse processes.

Definite integrals are tools for finding total quantities like area and volume. They appear frequently in physics, engineering, and statistics.

Integration Techniques

Integration techniques are strategies for finding antiderivatives or integrals of functions. These techniques are necessary because not all integrals can be solved using basic formulas like power rule or constant rule.
Integration by parts is one such technique that pairs functions strategically:

• Integration by Parts: Based on the product rule for differentiation, this technique splits integrals into more straightforward components.
• Choosing u and dv: Involves choosing which function in the integral becomes u and which becomes dv.
• Formula Application: Once components are selected, use the formula \(\int u \, dv = uv - \int v \, du\) to solve the integral.

Another set of techniques involves substitution and partial fraction decomposition, useful when dealing with polynomial or trigonometric functions. Selecting the right technique is key to simplifying and solving complex integrals.