Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_{-\pi /2}^{\pi /2}{\sin }^{4}x\left(1+\log \left(\frac{2+\sin x}{2-\sin x}\right)\right)dx$ is :

The integral provided is:

When you encounter limits of the form $[-\frac{\pi }{2},\frac{\pi }{2}]$, your mathematical intuition should immediately signal to check for symmetry. This is the key to unlocking the problem efficiently.

We start by splitting this intimidating integral into two manageable pieces. Let $I=I_1+I_2$, where:

By separating the terms, we have effectively isolated the complex logarithmic component from the simpler trigonometric power.

Now, let us focus on $I_2$. We define the integrand as $f_2(x)={\sin }^{4}x\log \left(\frac{2+\sin x}{2-\sin x}\right)$.

To check for parity, we evaluate $f_2(-x)$:

Since $\sin (-x)=-\sin x$ and the power of four makes the sine term positive, we obtain:

Notice that $\frac{2-\sin x}{2+\sin x}$ is the reciprocal of $\frac{2+\sin x}{2-\sin x}$. Using the property $\log (\frac{a}{b})=-\log (\frac{b}{a})$, we find that $f_2(-x)=-f_2(x)$.

This confirms that $f_2(x)$ is an odd function. The integral of an odd function over a symmetric interval $[-a,a]$ is always zero, as the area above the $x$-axis perfectly cancels the area below it. Thus, $I_2=0$.

We are left with $I=I_1={\int }_{-\pi /2}^{\pi /2}{\sin }^{4}xdx$. Since ${\sin }^{4}x$ is an even function, we apply the property ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$:

Now, we apply the Wallis Formula for ${\int }_0^{\pi /2}{\sin }^{n}xdx$ with $n=4$:

Finally, multiplying by the factor of two we extracted earlier:

The final result is $\frac{3\pi }{8}$. A seemingly impossible integral has been solved with elegance and symmetry.