If the value of the integral
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{x^2 \cos x}{1 + \pi^x} + \frac{1 + \sin^2 x}{1 + e^{\sin^x 2023}} \right) dx = \frac{\pi}{4} (\pi + a) - 2, \]
then the value of \(a\) is:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{x^2 \cos x}{1 + \pi^x} + \frac{1 + \sin^2 x}{1 + e^{\sin^{2023} x}} \right) dx \]
Notice that the integrand has symmetry properties, allowing us to split the integral and add:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{x^2 \cos x}{1 + \pi^x} + \frac{1 + \sin^2 x}{1 + e^{\sin^{2023}(-x)}} \right) dx \]
Adding the two integrals results in:
\[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( x^2 \cos x + 1 + \sin^2 x \right) dx \]
Solving this integral gives:
\[ I = \frac{\pi^2}{4} + \frac{3\pi}{4} - 2 \]
From the given equation, equate terms to find \(a = 3\).
So, the correct answer is: \(a = 3\)
Definite integral is an operation on functions which approximates the sum of the values (of the function) weighted by the length (or measure) of the intervals for which the function takes that value.
Definite integrals - Important Formulae Handbook
A real valued function being evaluated (integrated) over the closed interval [a, b] is written as :
\(\int_{a}^{b}f(x)dx\)
Definite integrals have a lot of applications. Its main application is that it is used to find out the area under the curve of a function, as shown below: