Understanding the Integral Problem

The question asks us to evaluate a specific indefinite integral involving trigonometric functions and then determine the values of the constants \(A\) and \(B\) by comparing the result to a given linear combination of \(x\) and \(\log |\sin(x-a)|\).

The given integral is \(\displaystyle\int\dfrac{\sin x}{\sin (x-a)}dx\).

The expected form of the result is \(Ax+B\log |sin(x-a)|+ C\), where \(A\), \(B\), and \(C\) are real constants.

Evaluating the Trigonometric Integral

We need to find the value of the integral \(\displaystyle\int\dfrac{\sin x}{\sin (x-a)}dx\). A common technique when the integrand is a ratio of trigonometric functions with shifted arguments is to manipulate the numerator so that it includes the argument of the denominator, \((x-a)\).

Step 1: Rewrite the Numerator using Trigonometric Identity

We can express the argument \(x\) in the numerator as \((x-a) + a\).

So, we have \(\sin x = \sin((x-a) + a)\).

Using the trigonometric sum identity for sine, \(\sin(P+Q) = \sin P \cos Q + \cos P \sin Q\), with \(P = x-a\) and \(Q = a\):
\(\sin x = \sin(x-a) \cos a + \cos(x-a) \sin a\)

Step 2: Substitute and Simplify the Integrand

Substitute this expression for \(\sin x\) back into the integral:

\(\displaystyle\int \frac{\sin(x-a) \cos a + \cos(x-a) \sin a}{\sin (x-a)} dx\)

Now, we can split the fraction into two terms:

\(\displaystyle\int \left( \frac{\sin(x-a) \cos a}{\sin (x-a)} + \frac{\cos(x-a) \sin a}{\sin (x-a)} \right) dx\)

Simplify the terms. The first term simplifies because \(\sin(x-a)\) cancels out. The second term uses the identity \(\frac{\cos \theta}{\sin \theta} = \cot \theta\).

\(\displaystyle\int \left( \cos a + \sin a \frac{\cos(x-a)}{\sin (x-a)} \right) dx = \int (\cos a + \sin a \cot(x-a)) dx\)

Step 3: Integrate Term by Term

Now, we integrate each term separately. Remember that \(a\) is a constant, so \(\cos a\) and \(\sin a\) are constants with respect to \(x\).

• The integral of the first term is \(\int \cos a \, dx\). Since \(\cos a\) is a constant, this is \(\cos a \int 1 \, dx = \cos a \cdot x\).
• The integral of the second term is \(\int \sin a \cot(x-a) \, dx\). Since \(\sin a\) is a constant, this is \(\sin a \int \cot(x-a) \, dx\).

The integral of \(\cot u\) with respect to \(u\) is \(\log |\sin u|\). Here, \(u = x-a\). The derivative of \(x-a\) with respect to \(x\) is 1, so the integral of \(\cot(x-a)\) is \(\log |\sin(x-a)|\).

So, \(\int \sin a \cot(x-a) \, dx = \sin a \log |\sin(x-a)| + C_1\), where \(C_1\) is the constant of integration for this part.

Combining the results from both terms, the complete integral is:

\(\displaystyle\int\dfrac{\sin x}{\sin (x-a)}dx = (\cos a) x + (\sin a) \log |\sin(x-a)| + C'\)

Here, \(C'\) represents the total constant of integration.

Comparing with the Given Form Ax + B log|sin(x-a)| + C

The given form of the integral result is \(Ax+B\log |sin(x-a)|+ C\).

We obtained the result \((\cos a) x + (\sin a) \log |\sin(x-a)| + C'\).

By comparing the coefficients of \(x\) and \(\log |\sin(x-a)|\), and the constant term, we can determine the values of \(A\) and \(B\).

• The coefficient of \(x\) in the given form is \(A\). In our result, it is \(\cos a\). Therefore, \(A = \cos a\).
• The coefficient of \(\log |\sin(x-a)|\) in the given form is \(B\). In our result, it is \(\sin a\). Therefore, \(B = \sin a\).
• The constant term is \(C\). In our result, it is the constant of integration \(C'\). The constant of integration can be any real number. Thus, \(C\) is an arbitrary real constant, \(C \in \mathbb{R}\).

Conclusion on Constants A, B, and C

Based on our evaluation and comparison, we find that \(A = \cos a\), \(B = \sin a\), and \(C \in \mathbb{R}\).

Let's examine the given options to find the one that matches our findings.

• Option 1: \(A = \sin a\), \(B \in \mathbb{R}\), \(C = \cos a\) - This does not match our derived values for A and B.
• Option 2: \(A = \cos a\), \(B = \sin a\), \(C \in \mathbb{R}\) - This perfectly matches our derived values for A and B, and the condition for C.
• Option 3: \(A \in \mathbb{R}\), \(B = \cos a\), \(C = \sin a\) - This does not match our derived values for A and B.
• Option 4: \(A = \sin a\), \(B = \cos a\), \(C \in \mathbb{R}\) - This does not match our derived values for A and B.

Therefore, the correct constants are \(A = \cos a\), \(B = \sin a\), and \(C \in \mathbb{R}\).

Revision Table: Key Concepts for Trigonometric Integral Evaluation

This table summarizes the essential concepts and steps used in solving this integral problem involving trigonometric functions.

Additional Information on Integral of Cotangent and Constants

Let's delve a bit deeper into the integral of the cotangent function, which was a key step in solving this problem.

The integral \(\int \cot u \, du\) is found by rewriting \(\cot u\) as \(\frac{\cos u}{\sin u}\) and using a substitution.

Let \(v = \sin u\). Then, the differential \(dv = \frac{d}{du}(\sin u) \, du = \cos u \, du\).

Substituting into the integral:

\(\displaystyle\int \cot u \, du = \int \frac{\cos u}{\sin u} \, du = \int \frac{1}{\sin u} (\cos u \, du)\)

With the substitution \(v = \sin u\) and \(dv = \cos u \, du\), the integral becomes:

\(\displaystyle\int \frac{1}{v} dv\)

This is a standard integral whose result is \(\log |v| + C_0\), where \(C_0\) is the constant of integration.

Substituting back \(v = \sin u\), we get:

\(\int \cot u \, du = \log |\sin u| + C_0\)

In our problem, we integrated \(\cot(x-a)\). Let \(u = x-a\). Then \(du = dx\). The integral is \(\int \cot u \, du\), which gives \(\log |\sin u| + C_1 = \log |\sin(x-a)| + C_1\).

Also, it's important to remember that the constant of integration \(C\) in an indefinite integral represents an entire family of antiderivatives. When comparing the result of an indefinite integral to a given form like \(Ax + B\log |\sin(x-a)|+ C\), the constant \(C\) is understood to be this arbitrary real constant.