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

The integral provided is:

At first glance, the denominator $1+a^x$ appears to defy standard integration techniques. However, in competitive mathematics, symmetric limits like $-\pi$ to $\pi$ almost always signal the presence of hidden symmetry.

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The King's Property is a powerful identity defined as:

In our case, the limits are $-\pi$ and $\pi$, so their sum $a+b=0$. Applying this property, we replace $x$ with $-x$:

Since $\cos (-x)=\cos x$, the numerator remains ${\cos }^{2}x$. The denominator transforms into $1+a^{-x}$.

We know that $a^{-x}=\frac{1}{a^x}$. Thus, the denominator becomes $1+\frac{1}{a^x}=\frac{a^x+1}{a^x}$.

Substituting this back into the integral, the $a^x$ term flips to the numerator:

Now, we add the original integral $I$ to the transformed integral $I$:

Because the denominators are identical, we combine the numerators:

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

Since ${\cos }^{2}x$ is an even function, we use the property ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$:

To solve this, we apply the half-angle identity ${\cos }^{2}x=\frac{1+\cos 2x}{2}$:

Evaluating at the limits, we obtain:

The final result is:

$$I=\frac{\pi }{2}$$