Key Concepts and Formulas

1. Symmetric Limits Property: For an integral $\int_{-a}^a f(x) dx$:
   • If $f(x)$ is odd ($f(-x) = -f(x)$), the integral is 0.
   • If $f(x)$ is even ($f(-x) = f(x)$), the integral is $2\int_0^a f(x) dx$.
2. King Property (Property 4): For an integral $\int_0^a f(x) dx$, we have $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.
3. Trigonometric Identities: $\sin(\pi - x) = \sin x$, $\cos(\pi - x) = -\cos x$.
4. Standard Integral: $\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$.

Step-by-Step Solution

Let the given integral be $I$. $I = \int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx$

Step 1: Decomposing the Integrand and Applying Symmetric Limit Properties

We can split the integrand into two terms: I = \int_{ - \pi }^\pi {\left( {{{2x} \over {1 + {{\cos }^2}x}}} + {{2x\sin x} \over {1 + {{\cos }^2}x}}} \right)} dx $I = \int_{ - \pi }^\pi {{{2x} \over {1 + {{\cos }^2}x}}} dx + \int_{ - \pi }^\pi {{{2x\sin x} \over {1 + {{\cos }^2}x}}} dx$

Let's analyze the parity of each integrand over the interval $[-\pi, \pi]$.

• For the first integral, let $f_1(x) = \frac{2x}{1 + \cos^2 x}$. $f_1(-x) = \frac{2(-x)}{1 + \cos^2(-x)} = \frac{-2x}{1 + \cos^2 x} = -f_1(x)$. Since $f_1(x)$ is an odd function, $\int_{ - \pi }^\pi {{{2x} \over {1 + {{\cos }^2}x}}} dx = 0$.

• For the second integral, let $f_2(x) = \frac{2x\sin x}{1 + \cos^2 x}$. $f_2(-x) = \frac{2(-x)\sin(-x)}{1 + \cos^2(-x)} = \frac{2(-x)(-\sin x)}{1 + \cos^2 x} = \frac{2x\sin x}{1 + \cos^2 x} = f_2(x)$. Since $f_2(x)$ is an even function, $\int_{ - \pi }^\pi {{{2x\sin x} \over {1 + {{\cos }^2}x}}} dx = 2\int_0^\pi {{{2x\sin x} \over {1 + {{\cos }^2}x}}} dx = 4\int_0^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}} dx$.

Thus, the integral $I$ simplifies to: $I = 0 + 4\int_0^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}} dx$ Let $I' = \int_0^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}} dx$, so $I = 4I'$.

Step 2: Applying the King Property

We apply the King Property ($\int_0^a f(x) dx = \int_0^a f(a-x) dx$) to $I'$ with $a = \pi$. $I' = \int_0^\pi {{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)} \over {1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx$ Using $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$ (so $\cos^2(\pi - x) = \cos^2 x$): $I' = \int_0^\pi {{{\left( {\pi - x} \right)\sin x} \over {1 + {{\cos }^2}x}}} dx$

Step 3: Combining the Integrals for $I'$

We have two expressions for $I'$:

1. $I' = \int_0^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}} dx$
2. $I' = \int_0^\pi {{{\left( {\pi - x} \right)\sin x} \over {1 + {{\cos }^2}x}}} dx$

Adding these two equations: $2I' = \int_0^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}} dx + \int_0^\pi {{{\left( {\pi - x} \right)\sin x} \over {1 + {{\cos }^2}x}}} dx$ $2I' = \int_0^\pi {\left( {{{x\sin x} \over {1 + {{\cos }^2}x}} + {{(\pi - x)\sin x} \over {1 + {{\cos }^2}x}}} \right)} dx$ $2I' = \int_0^\pi {{{\left( {x\sin x + \pi \sin x - x\sin x} \right)} \over {1 + {{\cos }^2}x}}} dx$ $2I' = \int_0^\pi {{{\pi \sin x} \over {1 + {{\cos }^2}x}}} dx$ $2I' = \pi \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx$ So, $I' = \frac{\pi}{2} \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx$.

Now, substituting this back into $I = 4I'$: $I = 4 \left( \frac{\pi}{2} \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx \right) = 2\pi \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx$

Step 4: Substitution and Evaluation

Let $t = \cos x$. Then $dt = -\sin x \, dx$, which means $\sin x \, dx = -dt$. We need to change the limits of integration:

• When $x = 0$, $t = \cos(0) = 1$.
• When $x = \pi$, $t = \cos(\pi) = -1$.

Substitute these into the integral: $I = 2\pi \int_1^{-1} {{1 \over {1 + {t^2}}}} (-dt)$ $I = -2\pi \int_1^{-1} {{1 \over {1 + {t^2}}}} dt$ Using the property $\int_a^b f(x) dx = -\int_b^a f(x) dx$, we reverse the limits: $I = 2\pi \int_{-1}^1 {{1 \over {1 + {t^2}}}} dt$

The integral of $\frac{1}{1+t^2}$ is $\tan^{-1}t$. $I = 2\pi \left[ {\tan^{ - 1}t} \right]_{-1}^1$ $I = 2\pi \left( {\tan^{ - 1}(1) - \tan^{ - 1}(-1)} \right)$ We know $\tan^{-1}(1) = \frac{\pi}{4}$ and $\tan^{-1}(-1) = -\frac{\pi}{4}$. $I = 2\pi \left( {\frac{\pi}{4} - \left( {-\frac{\pi}{4}} \right)} \right)$ $I = 2\pi \left( {\frac{\pi}{4} + \frac{\pi}{4}} \right)$ $I = 2\pi \left( {\frac{2\pi}{4}} \right)$ $I = 2\pi \left( {\frac{\pi}{2}} \right)$ $I = \pi^2$

Common Mistakes & Tips

• Incorrect Parity Analysis: Ensure that when checking for odd/even functions, you correctly apply the properties of trigonometric functions for negative arguments ($\sin(-x) = -\sin x$, $\cos(-x) = \cos x$).
• Algebraic Errors in Combining Integrals: Be meticulous when adding the two expressions for $I'$ after applying the King Property. The simplification of $x\sin x$ terms is key.
• Limit Changes in Substitution: Always remember to change the limits of integration when performing a substitution. Failure to do so will lead to an incorrect result.

Summary

The problem is solved by first splitting the integral and applying the symmetric limits property to eliminate one part of the integral by identifying it as an odd function. The remaining integral, which involves an even function, is simplified by applying the King Property. This allows for combining two forms of the integral, leading to a significant simplification. Finally, a trigonometric substitution is used to evaluate the resulting standard integral, yielding the final answer.

The final answer is $\boxed{{\pi^2}}$.