Integral Evaluation Steps

We need to evaluate the definite integral: $I = \int_{-\pi/6}^{\pi/6} \left( \frac{\pi + 4x^{11}}{1 - \sin(|x| + \pi/6)} \right) dx$

Analyzing the Integrand

Let the integrand be $f(x) = \frac{\pi + 4x^{11}}{1 - \sin(|x| + \pi/6)}$. The interval of integration $[-\pi/6, \pi/6]$ is symmetric around $x=0$. We examine the properties of $f(x)$.

Let $N(x) = \pi + 4x^{11}$ and $D(x) = 1 - \sin(|x| + \pi/6)$.

• The denominator $D(x)$ is an even function since $D(-x) = 1 - \sin(|-x| + \pi/6) = 1 - \sin(|x| + \pi/6) = D(x)$.
• The numerator $N(x)$ is neither odd nor even. $N(-x) = \pi + 4(-x)^{11} = \pi - 4x^{11}$.

We can decompose the integral using the property $\int_{-a}^{a} f(x) dx = \int_{-a}^{a} \frac{f(x) + f(-x)}{2} dx$. $f(x) + f(-x) = \frac{\pi + 4x^{11}}{D(x)} + \frac{\pi - 4x^{11}}{D(x)} = \frac{2\pi}{D(x)} = \frac{2\pi}{1 - \sin(|x| + \pi/6)}$. So, $I = \int_{-\pi/6}^{\pi/6} \frac{1}{2} \left( \frac{2\pi}{1 - \sin(|x| + \pi/6)} \right) dx = \int_{-\pi/6}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx$.

Simplifying the Integral

Let $g(x) = \frac{\pi}{1 - \sin(|x| + \pi/6)}$. Since $g(x)$ is an even function, we use the property $\int_{-a}^{a} g(x) dx = 2 \int_{0}^{a} g(x) dx$. $I = 2 \int_{0}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx$. In the interval $[0, \pi/6]$, $|x| = x$. $I = 2\pi \int_{0}^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$.

Substitution and Integration

Let $u = x + \pi/6$. Then $du = dx$. The limits of integration change: When $x=0$, $u = \pi/6$. When $x=\pi/6$, $u = \pi/3$. $I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1}{1 - \sin(u)} du$. To integrate $\frac{1}{1 - \sin(u)}$, multiply the numerator and denominator by $(1 + \sin(u))$: $\frac{1}{1 - \sin(u)} = \frac{1 + \sin(u)}{(1 - \sin(u))(1 + \sin(u))} = \frac{1 + \sin(u)}{1 - \sin^2(u)} = \frac{1 + \sin(u)}{\cos^2(u)}$ $= \frac{1}{\cos^2(u)} + \frac{\sin(u)}{\cos^2(u)} = \sec^2(u) + \sec(u)\tan(u)$. The integral becomes: $I = 2\pi \int_{\pi/6}^{\pi/3} (\sec^2(u) + \sec(u)\tan(u)) du$. The antiderivative is $\tan(u) + \sec(u)$. $I = 2\pi [\tan(u) + \sec(u)]_{\pi/6}^{\pi/3}$.

Applying Limits

Evaluate the antiderivative at the limits:

• At $u = \pi/3$: $\tan(\pi/3) + \sec(\pi/3) = \sqrt{3} + 2$.
• At $u = \pi/6$: $\tan(\pi/6) + \sec(\pi/6) = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.

Substitute these values back: