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Step-by-Step Detailed Solution
Step 1: Interpret the Integral and the Given Expression
We need to evaluate the integral
$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2 \cos x}{1+\pi^x} + \frac{1+\sin^2 x}{1+e^{\sin \left(x^{2123}\right)}}\right) \, dx $,
which is given to be equal to
$ \frac{\pi}{4}\left(\pi + a\right) - 2. $
We must find the value of $a$.
Step 2: Split the Integrand and Look for Symmetry
Define:
$ I_1(x) = \frac{x^2 \cos x}{1 + \pi^x} \quad \text{and} \quad I_2(x) = \frac{1 + \sin^2 x}{1 + e^{\sin\left(x^{2123}\right)}}. $
We will analyze each part separately under the transformation $x \mapsto -x$.
2.1 Symmetry of $I_1(x)$
Calculate $I_1(-x)$:
$$
I_1(-x)
= \frac{(-x)^2 \cos(-x)}{1 + \pi^{-x}}
= \frac{x^2 \cos x}{1 + \frac{1}{\pi^x}}
= \frac{x^2 \cos x \,\pi^x}{\pi^x + 1}.
$$
Adding $I_1(x)$ and $I_1(-x)$:
$$
I_1(x) + I_1(-x)
= \frac{x^2 \cos x}{1 + \pi^x} + \frac{x^2 \cos x \,\pi^x}{\pi^x + 1}
= x^2 \cos x.
$$
2.2 Symmetry of $I_2(x)$
Calculate $I_2(-x)$:
$$
I_2(-x)
= \frac{1 + \sin^2(-x)}{1 + e^{\sin\bigl((-x)^{2123}\bigr)}}
= \frac{1 + \sin^2 x}{1 + e^{-\sin(x^{2123})}}
= \frac{\bigl(1 + \sin^2 x\bigr)\, e^{\sin(x^{2123})}}{1 + e^{\sin(x^{2123})}}.
$$
Adding $I_2(x)$ and $I_2(-x)$:
$$
I_2(x) + I_2(-x)
= \frac{1 + \sin^2 x}{1 + e^{\sin(x^{2123})}}
+ \frac{\bigl(1 + \sin^2 x\bigr)\, e^{\sin(x^{2123})}}{1 + e^{\sin(x^{2123})}}
= 1 + \sin^2 x.
$$
Step 3: Express the Integral Using Symmetry
By the above symmetries:
$$
I_1(x) + I_1(-x) = x^2 \cos x,
\quad
I_2(x) + I_2(-x) = 1 + \sin^2 x.
$$
Hence,
$$
2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} I_1(x)\,dx
= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \,dx,
\quad
2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} I_2(x)\,dx
= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \bigl(1 + \sin^2 x\bigr)\,dx.
$$
Therefore, the original integral becomes:
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \Bigl(I_1(x) + I_2(x)\Bigr)\,dx
= \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x\,dx
+ \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \bigl(1 + \sin^2 x\bigr)\,dx.
$$
Step 4: Evaluate the Required Integrals
4.1 Integral of $x^2 \cos x$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$
$x^2 \cos x$ is an even function, so:
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx
= 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx.
$$
Use integration by parts on $ \int x^2 \cos x \, dx $:
$$
\int x^2 \cos x \, dx = x^2 \sin x - 2 \int x \sin x \, dx.
$$
Again, for $ \int x \sin x \, dx $, we can apply integration by parts or recall the standard result:
$$
\int x \sin x \, dx = -x \cos x + \sin x.
$$
Thus,
$$
\int x^2 \cos x \, dx
= x^2 \sin x + 2x \cos x - 2 \sin x + C.
$$
Evaluate from $0$ to $ \frac{\pi}{2} $:
$$
\int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx
= \Bigl[x^2 \sin x + 2x \cos x - 2\sin x\Bigr]_{0}^{\frac{\pi}{2}}
= \left(\frac{\pi^2}{4} \cdot 1 + \pi \cdot 0 - 2 \cdot 1\right) - 0
= \frac{\pi^2}{4} - 2.
$$
Hence,
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx
= 2 \Bigl(\frac{\pi^2}{4} - 2\Bigr)
= \frac{\pi^2}{2} - 4.
$$
4.2 Integral of $1 + \sin^2 x$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$
$1 + \sin^2 x$ is also even, so:
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \bigl(1 + \sin^2 x\bigr)\,dx
= 2 \int_{0}^{\frac{\pi}{2}} \bigl(1 + \sin^2 x\bigr)\,dx.
$$
We know:
$$
\int_{0}^{\frac{\pi}{2}} 1\,dx = \frac{\pi}{2},
\quad
\int_{0}^{\frac{\pi}{2}} \sin^2 x\,dx = \frac{\pi}{4}.
$$
Therefore,
$$
\int_{0}^{\frac{\pi}{2}} (1 + \sin^2 x)\,dx
= \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}.
$$
Doubling this:
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \sin^2 x)\,dx
= 2 \cdot \frac{3\pi}{4} = \frac{3\pi}{2}.
$$
Step 5: Combine the Results
Recall:
$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \Bigl(I_1(x) + I_2(x)\Bigr)\,dx
= \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x\,dx
+ \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \bigl(1 + \sin^2 x\bigr)\,dx.
$$
Substitute the evaluated integrals:
$$
= \frac{1}{2} \Bigl(\frac{\pi^2}{2} - 4\Bigr)
+ \frac{1}{2} \Bigl(\frac{3\pi}{2}\Bigr)
= \frac{\pi^2}{4} - 2 + \frac{3\pi}{4}.
$$
Rearrange:
$$
= \frac{\pi}{4}(\pi + 3) - 2.
$$
Step 6: Identify the Value of $a$
According to the problem statement, the integral equals
$ \frac{\pi}{4} (\pi + a) - 2. $
By comparing
$ \frac{\pi}{4}(\pi + 3) - 2 $
with
$ \frac{\pi}{4}(\pi + a) - 2, $
we conclude
$ a = 3. $
Final Answer
The value of $a$ is 3.
