Animated Solution for Mathematics - Definite Integration: The value of ${\int }_{-\pi }^{\pi }\frac{{\cos }^{2}x}{1+a^x}dx,a>0$, is

Whenever you encounter an integral with limits from $-a$ to $a$, your brain should immediately light up. This is not a coincidence; it is a massive hint. We are looking at a symmetric interval for the integral:

The function we are integrating is $f(x)=\frac{{\cos }^{2}x}{1+a^x}$. The numerator, ${\cos }^{2}x$, is a well-behaved even function, while the denominator, $1+a^x$, is the "chaos factor" that is neither even nor odd.

We utilize the 'King's Property' of definite integrals, which states:

Since our limits are $-\pi$ and $\pi$, their sum is $0$. Applying this property, our integral $I$ is also equal to ${\int }_{-\pi }^{\pi }f(-x)dx$.

Substituting $-x$ into the function, the numerator remains ${\cos }^2(-x)={\cos }^{2}x$ because cosine is an even function. The denominator becomes:

Now, let us look at our transformed integral:

We now have two versions of the same integral $I$. Adding them together, we get:

Combining the numerators over the common denominator:

The term $(1+a^x)$ cancels out completely, leaving us with:

Since ${\cos }^{2}x$ is an even function, we can simplify the integral to $2I=2{\int }_0^{\pi }{\cos }^{2}xdx$, which implies:

Using the power-reduction identity ${\cos }^{2}x=\frac{1+\cos 2x}{2}$, we proceed:

Evaluating at the limits:

The final result is $\frac{\pi }{2}$.