Problem 39 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \cot 2 u d u=\) (A) \(\ln |\sin u|+C\) (B) \(\frac{1}{2} \ln |\sin 2 u|+C\) (C) \(-\frac{1}{2} \csc ^{2} 2 u+C\) (D) \(2 \ln |\sin 2 u|+C\)

Step-by-Step Solution

Answer

Option (B): \(\frac{1}{2} \ln |\sin 2u| + C\).

Step 1: Recall the Cotangent Definition

The cotangent function, \(\cot u\), is the reciprocal of the tangent and can be expressed as \(\cot u = \frac{\cos u}{\sin u}\). This is important for substituting in integrals.

Step 2: Transform the Integral

We need to integrate \(\int \cot 2u \, du\). First, express \(\cot 2u\) as \(\frac{\cos 2u}{\sin 2u}\). The integral becomes \(\int \frac{\cos 2u}{\sin 2u} \, du\).

Step 3: Apply Substitution

Let \(v = \sin 2u\), then \(dv = 2\cos 2u \, du\), so \(du = \frac{dv}{2\cos 2u}\). The integral becomes: \(\int \frac{1}{2} \frac{1}{v} \, dv\).

Step 4: Integrate the Transformed Function

Now, integrate \(\int \frac{1}{2} \frac{1}{v} \, dv\), which simplifies to \(\frac{1}{2} \ln |v| + C\). This substitution leads to \(\frac{1}{2} \ln |\sin 2u| + C\), reverting \(v\) back to \(\sin 2u\).

Step 5: Confirm the Result Against Options

We've derived \(\frac{1}{2} \ln |\sin 2u| + C\). Comparing with given options, this matches option (B).

Key Concepts

Trigonometric Integrals

Trigonometric integrals involve integration of functions composed of trigonometric functions like sine, cosine, tangent, or cotangent. These integrals often require manipulation and substitution to simplify the expressions. For example, integrating functions involving cotangent, like in our exercise, uses the identity \( \cot u = \frac{\cos u}{\sin u} \). This transforms the integral involving cotangent into a more manageable form using sine and cosine.

• In our example, \( \int \cot 2u \, du \) was transformed into \( \int \frac{\cos 2u}{\sin 2u} \, du \).
• This setup is more straightforward because it allows us to utilize substitution methods effectively.

Trigonometric identities are a crucial tool in simplifying and solving integrals, helping to reveal substitution opportunities and make the integration process easier to handle.

Integration by Substitution

Integration by substitution is a powerful method used to simplify the integration process. It is akin to reversing the chain rule in differentiation and helps in transforming complex integrals into simpler ones. In our example, we used substitution to handle the integral \( \int \frac{\cos 2u}{\sin 2u} \, du \).

• We defined a new variable \( v \) as \( \sin 2u \), which simplified the integral to \( \int \frac{1}{2} \frac{1}{v} \, dv \).
• This dramatically simplifies the process as it allows us to integrate using the natural logarithm rule for \( \frac{1}{v} \), leading to \( \frac{1}{2} \ln |v| + C \).

By solving this modified integral, substitution allows us to navigate through potentially complex trigonometric expressions and focus on more simplified algebraic forms.

Definite and Indefinite Integrals

In calculus, integrals can be either definite or indefinite. Indefinite integrals, like the one in our exercise, do not have specified limits and thus contain a constant of integration, \( C \), reflecting all possible antiderivatives of a function.

• An indefinite integral is represented as \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative of \( f(x) \).
• In our exercise, after substitution and integration, we encountered \( \frac{1}{2} \ln |\sin 2u| + C \), an indefinite integral.

Definite integrals, on the other hand, are evaluated over specific intervals and result in a numerical value rather than a function. For completion, it is always essential in indefinite integrals to include the constant \( C \), as it accounts for any vertical shift in the antiderivative function, representing the family of all possible solutions.