1. Key Concepts and Formulas

• Trigonometric Identities: Conversion of $\cot x$ and $\csc x$ to $\sin x$ and $\cos x$ is crucial for simplification.
• Definite Integration Properties: The property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ is often useful for definite integrals with symmetric limits.
• Standard Integrals: Knowledge of the integral of $\sec^2 x$ and related forms.

2. Step-by-Step Solution

Let the given integral be $I$. $I = \int\limits_0^{{\pi \over 2}} {{{\cot x} \over {\cot x + \cos ecx}}} dx$

Step 1: Simplify the Integrand

To simplify the integrand, we express $\cot x$ and $\csc x$ in terms of $\sin x$ and $\cos x$. We know that $\cot x = \frac{\cos x}{\sin x}$ and $\csc x = \frac{1}{\sin x}$.

Substituting these into the integrand: $\frac{\cot x}{\cot x + \csc x} = \frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} + \frac{1}{\sin x}}$ Multiply the numerator and denominator by $\sin x$ to clear the complex fraction: $= \frac{\left(\frac{\cos x}{\sin x}\right) \cdot \sin x}{\left(\frac{\cos x}{\sin x} + \frac{1}{\sin x}\right) \cdot \sin x} = \frac{\cos x}{\cos x + 1}$ Thus, the integral becomes: $I = \int\limits_0^{{\pi \over 2}} {\frac{\cos x}{1 + \cos x}} dx$

Step 2: Apply the Property of Definite Integrals

We can use the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here, $a=0$ and $b=\frac{\pi}{2}$. Let $x' = a+b-x = 0 + \frac{\pi}{2} - x = \frac{\pi}{2} - x$. Then $dx' = -dx$. When $x=0$, $x' = \frac{\pi}{2}$. When $x=\frac{\pi}{2}$, $x'=0$. Also, $\cos x = \cos(\frac{\pi}{2} - x') = \sin x'$. The integral becomes: $I = \int\limits_{\pi/2}^0 {\frac{\sin x'}{1 + \sin x'}} (-dx')$ $I = \int\limits_0^{\pi/2} {\frac{\sin x'}{1 + \sin x'}} dx'$ Replacing the dummy variable $x'$ with $x$: $I = \int\limits_0^{\pi/2} {\frac{\sin x}{1 + \sin x}} dx$

Step 3: Combine the Two Forms of the Integral

Now we have two expressions for $I$: $I = \int\limits_0^{{\pi \over 2}} {\frac{\cos x}{1 + \cos x}} dx \quad (*)$ $I = \int\limits_0^{\pi/2} {\frac{\sin x}{1 + \sin x}} dx \quad (**)$ Adding $(*)$ and $(**)$: $2I = \int\limits_0^{\pi/2} {\frac{\cos x}{1 + \cos x}} dx + \int\limits_0^{\pi/2} {\frac{\sin x}{1 + \sin x}} dx$ $2I = \int\limits_0^{\pi/2} {\left(\frac{\cos x}{1 + \cos x} + \frac{\sin x}{1 + \sin x}\right)} dx$ $2I = \int\limits_0^{\pi/2} {\frac{\cos x(1 + \sin x) + \sin x(1 + \cos x)}{(1 + \cos x)(1 + \sin x)}} dx$ $2I = \int\limits_0^{\pi/2} {\frac{\cos x + \cos x \sin x + \sin x + \sin x \cos x}{(1 + \cos x)(1 + \sin x)}} dx$ $2I = \int\limits_0^{\pi/2} {\frac{\cos x + \sin x + 2 \sin x \cos x}{(1 + \cos x)(1 + \sin x)}} dx$ This approach seems complicated. Let's reconsider the simplification in Step 1.

Alternative Step 1 (Revisited): Simplify the Integrand further

We had: $I = \int\limits_0^{{\pi \over 2}} {\frac{\cos x}{1 + \cos x}} dx$ We can rewrite the integrand by adding and subtracting 1 in the numerator: $\frac{\cos x}{1 + \cos x} = \frac{1 + \cos x - 1}{1 + \cos x} = 1 - \frac{1}{1 + \cos x}$ Now, use the half-angle identity for cosine: $1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$. $\frac{1}{1 + \cos x} = \frac{1}{2\cos^2\left(\frac{x}{2}\right)} = \frac{1}{2} \sec^2\left(\frac{x}{2}\right)$ So the integrand is: $1 - \frac{1}{2} \sec^2\left(\frac{x}{2}\right)$

Step 2 (Revisited): Integrate the Simplified Expression

Now, we integrate this simplified expression: $I = \int\limits_0^{{\pi \over 2}} {\left(1 - \frac{1}{2} \sec^2\left(\frac{x}{2}\right)\right)} dx$ $I = \int\limits_0^{{\pi \over 2}} 1 \, dx - \int\limits_0^{{\pi \over 2}} \frac{1}{2} \sec^2\left(\frac{x}{2}\right) dx$

The first part is straightforward: $\int\limits_0^{{\pi \over 2}} 1 \, dx = [x]_0^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$

For the second part, we use the integration formula $\int \sec^2(ax) dx = \frac{1}{a}\tan(ax)$. Here $a = \frac{1}{2}$. $\int\limits_0^{{\pi \over 2}} \frac{1}{2} \sec^2\left(\frac{x}{2}\right) dx = \frac{1}{2} \int\limits_0^{{\pi \over 2}} \sec^2\left(\frac{x}{2}\right) dx$ Let $u = \frac{x}{2}$, so $du = \frac{1}{2} dx$, which means $dx = 2 du$. When $x=0$, $u=0$. When $x=\frac{\pi}{2}$, $u=\frac{\pi}{4}$. $\frac{1}{2} \int\limits_0^{\pi/4} \sec^2(u) (2 du) = \int\limits_0^{\pi/4} \sec^2(u) du$ $= [\tan u]_0^{\pi/4} = \tan\left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1$

So, combining the two parts: $I = \frac{\pi}{2} - 1$

Step 3: Match the Result with the Given Form

The problem states that the integral is equal to $m(\pi + n)$. We have $I = \frac{\pi}{2} - 1$. We need to rewrite this in the form $m(\pi + n)$. $I = \frac{1}{2}(\pi - 2)$ This is not exactly in the form $m(\pi + n)$. Let's re-examine the problem statement and our result. The given form is $m(\pi + n)$. Our result is $I = \frac{\pi}{2} - 1$.

Let's try to match it differently. We have $I = \frac{\pi}{2} - 1$. We can write this as: $I = m\pi + mn$.

Comparing $\frac{\pi}{2} - 1$ with $m\pi + mn$: We can see that $m = \frac{1}{2}$. Then, $mn = -1$. Substituting $m = \frac{1}{2}$ into $mn = -1$: $\frac{1}{2}n = -1$ $n = -2$.

So, $m = \frac{1}{2}$ and $n = -2$. The integral is $I = \frac{1}{2}(\pi + (-2)) = \frac{1}{2}(\pi - 2)$.

The question asks for the value of $m \cdot n$. $m \cdot n = \left(\frac{1}{2}\right) \cdot (-2) = -1$.

Let's double-check the original form given in the question: $m(\pi + n)$. Our calculated integral is $\frac{\pi}{2} - 1$. We need to express $\frac{\pi}{2} - 1$ in the form $m(\pi + n)$.

If we set $m = \frac{1}{2}$, then we have $\frac{1}{2}(\pi + n)$. $\frac{1}{2}(\pi + n) = \frac{\pi}{2} + \frac{n}{2}$. Comparing this with $\frac{\pi}{2} - 1$, we get: $\frac{n}{2} = -1 \implies n = -2$. So, $m = \frac{1}{2}$ and $n = -2$. Then $m \cdot n = \frac{1}{2} \cdot (-2) = -1$.

This matches option (A).

3. Common Mistakes & Tips

• Algebraic Errors: Be extremely careful when simplifying fractions involving trigonometric functions.
• Integration of $\sec^2(ax)$: Remember the factor of $\frac{1}{a}$ when integrating $\sec^2(ax)$.
• Matching Forms: When a result needs to be matched with a specific algebraic form like $m(\pi + n)$, ensure all terms are correctly identified.

4. Summary

The problem involves evaluating a definite integral of a trigonometric function. First, the integrand was simplified by converting $\cot x$ and $\csc x$ into their $\sin x$ and $\cos x$ forms, leading to $\frac{\cos x}{1+\cos x}$. This was further rewritten as $1 - \frac{1}{1+\cos x}$. Using the half-angle identity for $\cos x$, the integral was transformed into $\int_0^{\pi/2} (1 - \frac{1}{2}\sec^2(\frac{x}{2})) dx$. Integrating this expression yielded $\frac{\pi}{2} - 1$. This result was then equated to the given form $m(\pi + n)$, which allowed us to determine $m = \frac{1}{2}$ and $n = -2$. The product $m \cdot n$ was calculated to be $-1$.

The final answer is $\boxed{-1}$.