Animated Solution for Mathematics - Definite Integration: The value of ${\int }_{-\pi /2}^{\pi /2}\left(\frac{1+{\sin }^{2}x}{1+{\pi }^{\sin x}}\right)dx$ is

We begin by defining our integral as:

The moment you see symmetric limits from $-\frac{\pi }{2}$ to $\frac{\pi }{2}$, your mathematical spider-sense should tingle. This is the hallmark of the King's Property, one of the most powerful tools in your arsenal.

The property states that:

In our case, $a=-\frac{\pi }{2}$ and $b=\frac{\pi }{2}$, so $a+b=0$. This means we are simply replacing $x$ with $-x$.

When we substitute $x\to -x$, our integral becomes:

Recall your trigonometry: $\sin (-x)=-\sin x$, and because of the square, ${\sin }^2(-x)={\sin }^{2}x$. The numerator remains unchanged, but the denominator becomes $1+{\pi }^{-\sin x}$.

Since ${\pi }^{-\sin x}=\frac{1}{{\pi }^{\sin x}}$, we can simplify the denominator to $\frac{{\pi }^{\sin x}+1}{{\pi }^{\sin x}}$. Flipping the fraction, our new integral becomes:

Now, we add our two expressions for $I$ together:

Notice that both terms share the same denominator. Combining the numerators and factoring out $(1+{\sin }^{2}x)$, we get:

The exponential terms cancel out perfectly, leaving us with:

Since $1+{\sin }^{2}x$ is an even function, we use the symmetry property ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$:

Using the identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$, we substitute:

Integrating term by term, we evaluate:

Plugging in the limits, the upper limit yields $\frac{3}{2}(\frac{\pi }{2})-\frac{1}{4}\sin (\pi )=\frac{3\pi }{4}-0$. The lower limit is $0$.

The final answer is $\frac{3\pi }{4}$.