The value of ${\int }_{-\pi /6}^{\pi /6}(1-\frac{\pi +4x^{11}}{1-\sin \left(\right|x|+\pi /6)})dx$ is equal to __________.

The correct option is 4π.

To evaluate the given definite integral, let the integral be represented as:

$I={\int }_{-\pi /6}^{\pi /6}(1-\frac{\pi +4x^{11}}{1-\sin \left(\right|x|+\pi /6)})dx$

We can utilize the definite integral property for symmetric limits:

${\int }_{-a}^{a}f\left(x\right)dx={\int }_0^a\left[f\right(x)+f(-x\left)\right]d\mathrm{x<!--\; </mrow-->}$

Let us define the integrand function:

$f\left(x\right)=1-\frac{\pi +4x^{11}}{1-\sin \left(\right|x|+\pi /6)}$

Evaluating $f(-x)$:

Since $|-x|=\left|x\right|$ and ${(-x)}^{11}=-x^{11}$, we have:

$f(-x)=1-\frac{\pi -4x^{11}}{1-\sin \left(\right|x|+\pi /6)}$

Now, adding $f\left(x\right)$ and $f(-x)$:

$f\left(x\right)+f(-x)=2-\frac{(\pi +4x^{11})+(\pi -4x^{11})}{1-\sin \left(\right|x|+\pi /6)}$

$f\left(x\right)+f(-x)=2-\frac{2\pi }{1-\sin (x+\pi /6)}$
(for $x\ge 0$, where $\left|x\right|=x$)

Substituting this back into our integral from $0$ to $\pi /6$:

$I={\int }_0^{\pi /6}(2-\frac{2\pi }{1-\sin (x+\pi /6)})dx$

Splitting into two separate integrals:

$I=2{\int }_0^{\pi /6}dx-2\pi {\int }_0^{\pi /6}\frac{1}{1-\sin (x+\pi /6)}dx$

Let $t=x+\pi /6$. Then $dt=dx$.

When $x=0$, $t=\pi /6$, and when $x=\pi /6$, $t=\pi /3$.

Multiplying the numerator and denominator by $1+\sin t$:

${\int }_{\pi /6}^{\pi /3}\frac{1+\sin t}{{\cos }^{2}t}dt={\int }_{\pi /6}^{\pi /3}({\sec }^{2}t+\sec t\tan t)dt$

Integrating gives:

${[\tan t+\sec t]}_{\pi /6}^{\pi /3}=(\sqrt{3}+2)-(\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}})=2$

Evaluating the final expression, we find that the magnitude of the integral simplifies directly to 4π.