Analyzing and Comparing Definite Integrals

The question asks us to compare the values of two definite integrals, A and B, defined over the interval <strong>[0, π]</strong>.

The given integrals are:

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

To find the relationship between these integrals, we can use a standard property of definite integrals. The property states that for a continuous function \(f(x)\) over the interval <strong>[0, a]</strong>, the following holds:

\(\mathop \smallint \limits_0^a f(x) {\rm{dx}} = \mathop \smallint \limits_0^a f(a - x) {\rm{dx}}\)

Let's apply this property to Integral A. Here, \(a = \pi\).

\({\rm{A}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{\sin {\rm{x}}}}{{\sin {\rm{x}} + \cos {\rm{x}}}} {\rm{dx}}\)

Using the property, we replace \(x\) with \((\pi - x)\) in the integrand:

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

Now, we use the trigonometric identities for angles involving <strong>π</strong>:

• \(\sin (\pi - x) = \sin x\)
• \(\cos (\pi - x) = -\cos x\)

Substituting these identities into the integral expression for A:

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

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

Let's look closely at the resulting integral:

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

This is exactly the definition of Integral B.

So, by applying the property \(\mathop \smallint \limits_0^a f(x) {\rm{dx}} = \mathop \smallint \limits_0^a f(a - x) {\rm{dx}}\) to Integral A, we found that A is equal to B.

Therefore, the relationship between A and B is <strong>A = B</strong>.

Comparison of Integral A and Integral B

We started with:

\({\rm{A}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{\sin {\rm{x\;dx}}}}{{\sin {\rm{x}} + \cos {\rm{x}}}}\) (Equation 1)

Applying the property \(\int_0^\pi f(x) dx = \int_0^\pi f(\pi - x) dx\) gives:

\({\rm{A}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{\sin (\pi - x){\rm{dx}}}}{{\sin (\pi - x) + \cos (\pi - x)}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{\sin {\rm{x\;dx}}}}{{\sin {\rm{x}} - \cos {\rm{x}}}}\) (Equation 2)

We also have the definition of B:

\({\rm{B}} = \mathop \smallint \limits_0^{\rm{\pi }} \frac{{\sin {\rm{x\;dx}}}}{{\sin {\rm{x}} - \cos {\rm{x}}}}\) (Equation 3)

Comparing Equation 2 and Equation 3, we clearly see that \({\rm{A}} = {\rm{B}}\).

Conclusion on the Relationship

Based on the application of the definite integral property, we have established that Integral A and Integral B are equal.

Let's check the given options:

Our derived relationship <strong>A = B</strong> matches Option 3.

Revision Table: Key Concepts

Additional Information: Definite Integral Properties

Definite integrals have several useful properties that can simplify evaluation or help establish relationships between integrals without direct computation. Besides the property used in this problem, here are a few others:

• Linearity: \(\mathop \smallint \limits_a^b [cf(x) + dg(x)] dx = c\mathop \smallint \limits_a^b f(x) dx + d\mathop \smallint \limits_a^b g(x) dx\)
• Interval Additivity: \(\mathop \smallint \limits_a^b f(x) dx = \mathop \smallint \limits_a^c f(x) dx + \mathop \smallint \limits_c^b f(x) dx\), where \(a < c < b\).
• Changing Limits: \(\mathop \smallint \limits_a^b f(x) dx = -\mathop \smallint \limits_b^a f(x) dx\)
• Property \(\int_0^{2a} f(x) dx = \int_0^a f(x) dx + \int_0^a f(2a-x) dx\). This can be further simplified if \(f(2a-x) = f(x)\) or \(f(2a-x) = -f(x)\).

Understanding these properties is essential for solving various problems involving definite integrals efficiently during exam preparation.