Evaluating Definite Trigonometric Integral B

The problem asks for the value of the definite integral \({\rm{B}}\), defined as:

\({\rm{B}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{\sin {\rm{x\;dx}}}}{{\sin {\rm{x}} - \cos {\rm{x}}}}\)

This is a definite integral involving trigonometric functions over the interval \([0, \pi]\). First, let's analyze the integrand \(\frac{{\sin {\rm{x}}}}{{\sin {\rm{x}} - \cos {\rm{x}}}}\). The denominator, \(\sin {\rm{x}} - \cos {\rm{x}}\), becomes zero when \(\sin {\rm{x}} = \cos {\rm{x}}\). In the interval \([0, \pi]\), this occurs at \({\rm{x}} = \frac{{\rm{\pi }}}{4}\). Since the discontinuity at \({\rm{x}} = \frac{{\rm{\pi }}}{4}\) is within the interval of integration, this integral is technically an improper integral.

However, given the multiple-choice options are finite values, the problem likely intends for a formal evaluation using the Fundamental Theorem of Calculus after finding the indefinite integral, often overlooking the discontinuity in this context. Let's proceed with this approach.

We need to find the indefinite integral \(\mathop \smallint \frac{{\sin {\rm{x\;dx}}}}{{\sin {\rm{x}} - \cos {\rm{x}}}}\). Let the integrand be \({\rm{f}}({\rm{x}}) = \frac{{\sin {\rm{x}}}}{{\sin {\rm{x}} - \cos {\rm{x}}}}\). We can try to write the numerator (\(\sin {\rm{x}}\)) as a linear combination of the denominator (\(\sin {\rm{x}} - \cos {\rm{x}}\)) and its derivative (\(\cos {\rm{x}} - (-\sin {\rm{x}}) = \cos {\rm{x}} + \sin {\rm{x}}\)).

Let \(\sin {\rm{x}} = {\rm{A}}(\sin {\rm{x}} - \cos {\rm{x}}) + {\rm{B}}(\cos {\rm{x}} + \sin {\rm{x}})\).

Expanding this equation:

\(\sin {\rm{x}} = {\rm{A}}\sin {\rm{x}} - {\rm{A}}\cos {\rm{x}} + {\rm{B}}\cos {\rm{x}} + {\rm{B}}\sin {\rm{x}}\)

\(\sin {\rm{x}} = ({\rm{A}} + {\rm{B}})\sin {\rm{x}} + ({\rm{B}} - {\rm{A}})\cos {\rm{x}}\)

Comparing the coefficients of \(\sin {\rm{x}}\) and \(\cos {\rm{x}}\) on both sides:

• Coefficient of \(\sin {\rm{x}}\): \({\rm{A}} + {\rm{B}} = 1\)
• Coefficient of \(\cos {\rm{x}}\): \({\rm{B}} - {\rm{A}} = 0\)

From the second equation, \({\rm{B}} = {\rm{A}}\). Substituting this into the first equation, we get \({\rm{A}} + {\rm{A}} = 1\), which means \(2{\rm{A}} = 1\), so \({\rm{A}} = \frac{1}{2}\). Since \({\rm{B}} = {\rm{A}}\), we also have \({\rm{B}} = \frac{1}{2}\).

Thus, we can rewrite the numerator as \(\sin {\rm{x}} = \frac{1}{2}(\sin {\rm{x}} - \cos {\rm{x}}) + \frac{1}{2}(\cos {\rm{x}} + \sin {\rm{x}})\).

Now, substitute this back into the integral:

\(\mathop \smallint \frac{\frac{1}{2}(\sin {\rm{x}} - \cos {\rm{x}}) + \frac{1}{2}(\cos {\rm{x}} + \sin {\rm{x}})}{{\sin {\rm{x}} - \cos {\rm{x}}}} {\rm{dx}}\)

\(= \mathop \smallint \left( \frac{1}{2} + \frac{\frac{1}{2}(\cos {\rm{x}} + \sin {\rm{x}})}{{\sin {\rm{x}} - \cos {\rm{x}}}} \right) {\rm{dx}}\)

\(= \frac{1}{2}\mathop \smallint 1 {\rm{dx}} + \frac{1}{2}\mathop \smallint \frac{{\cos {\rm{x}} + \sin {\rm{x}}}}{{\sin {\rm{x}} - \cos {\rm{x}}}} {\rm{dx}}\)

The first integral is simply \(\frac{1}{2}{\rm{x}}\). For the second integral, let \({\rm{u}} = \sin {\rm{x}} - \cos {\rm{x}}\). Then, the differential \({\rm{du}} = (\cos {\rm{x}} - (-\sin {\rm{x}})) {\rm{dx}} = (\cos {\rm{x}} + \sin {\rm{x}}) {\rm{dx}}\). The integral becomes:

\(\frac{1}{2}\mathop \smallint \frac{1}{{\rm{u}}} {\rm{du}} = \frac{1}{2}\ln |{\rm{u}}| + {\rm{C}} = \frac{1}{2}\ln |\sin {\rm{x}} - \cos {\rm{x}}| + {\rm{C}}\)

So, the indefinite integral is \(\frac{1}{2}{\rm{x}} + \frac{1}{2}\ln |\sin {\rm{x}} - \cos {\rm{x}}| + {\rm{C}}\).

Now, we evaluate the definite integral \({\rm{B}}\) by applying the limits of integration from 0 to \(\pi\):

\({\rm{B}} = \left[ \frac{1}{2}{\rm{x}} + \frac{1}{2}\ln |\sin {\rm{x}} - \cos {\rm{x}}| \right]_0^{\rm{\pi }}\)

\({\rm{B}} = \left( \frac{1}{2}(\pi) + \frac{1}{2}\ln |\sin(\pi) - \cos(\pi)| \right) - \left( \frac{1}{2}(0) + \frac{1}{2}\ln |\sin(0) - \cos(0)| \right)\)

Evaluate the trigonometric functions at the limits:

• \(\sin(\pi) = 0\)
• \(\cos(\pi) = -1\)
• \(\sin(0) = 0\)
• \(\cos(0) = 1\)

Substitute these values:

\({\rm{B}} = \left( \frac{\pi}{2} + \frac{1}{2}\ln |0 - (-1)| \right) - \left( 0 + \frac{1}{2}\ln |0 - 1| \right)\)

\({\rm{B}} = \left( \frac{\pi}{2} + \frac{1}{2}\ln |1| \right) - \left( \frac{1}{2}\ln |-1| \right)\)

Since \(\ln(1) = 0\) and \(\ln |-1| = \ln(1) = 0\):

\({\rm{B}} = \left( \frac{\pi}{2} + \frac{1}{2}(0) \right) - \left( \frac{1}{2}(0) \right)\)

\({\rm{B}} = \frac{\pi}{2} - 0 = \frac{\pi}{2}\)

Thus, the value of integral B is \(\frac{{\rm{\pi }}}{2}\).

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Additional Information on Definite Integrals and Discontinuities

A definite integral \(\mathop \smallint \limits_a^b f(x) \text{d}x\) is formally defined for a continuous function \(f(x)\) on the interval \([a, b]\). If the function \(f(x)\) has a discontinuity within the interval \((a, b)\), the integral is classified as an improper integral. For an improper integral to converge to a finite value, the limits of the integrals around the point of discontinuity must exist and be finite. In the case of integral B, the integrand has a discontinuity at \(x = \frac{\pi}{4}\).

Evaluating the integral as \(\lim_{\epsilon_1 \to 0^+} \mathop \smallint \limits_0^{\frac{\pi}{4}-\epsilon_1} f(x) \text{d}x + \lim_{\epsilon_2 \to 0^+} \mathop \smallint \limits_{\frac{\pi}{4}+\epsilon_2}^{\pi} f(x) \text{d}x\) would typically involve evaluating limits of the form \(\ln(|\text{something approaching } 0|)\), which tends towards \(-\infty\). Therefore, standard rigorous evaluation would show that the integral diverges.

However, in many applied contexts or specific types of problems (like this one appears to be, given the finite options), a formal application of the Fundamental Theorem of Calculus is expected, treating the integral as if the antiderivative is valid across the interval boundaries, especially if the discontinuity is removable in a specific sense or a principal value is considered. The method shown above reflects this common approach when a finite answer is anticipated from such a problem structure in multiple-choice tests.

The value of integral B is found to be \(\frac{{\rm{\pi }}}{2}\) using this formal evaluation method.