Question

The value of \(\int_{-\pi/6}^{\pi/6} \frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6) dx\) is equal to:}

• A. \(8\pi\)
• B. \(2\pi\)
• C. \(6\pi\)
• D. \(4\pi\)

Hint:

When faced with a definite integral on a symmetric interval like \([-a, a]\), always check if the integrand is even, odd, or can be split into even and odd parts.
This simple check can often simplify the problem significantly, as the integral of the odd part will be zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to evaluate a definite integral over a symmetric interval \([-\pi/6, \pi/6]\). The integrand has a complicated form, which suggests using properties of definite integrals, such as even and odd functions.
Step 2: Key Formula or Approach:
1. Splitting the integral: \(\int (f(x)+g(x)) dx = \int f(x) dx + \int g(x) dx\).
2. Even/Odd function property: For a symmetric interval \([-a, a]\):
- If \(f(x)\) is an odd function (\(f(-x) = -f(x)\)), then \(\int_{-a}^{a} f(x) dx = 0\).
- If \(f(x)\) is an even function (\(f(-x) = f(x)\)), then \(\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx\).
3. Trigonometric identity: \(1 - \sin(\theta) = 1 - \cos(\pi/2 - \theta) = 2\sin^2(\frac{\pi/2-\theta}{2}) = 2\sin^2(\pi/4-\theta/2)\).
4. Standard integral: \(\int \csc^2(u) du = -\cot(u) + C\).
Step 3: Detailed Explanation:
Let the given integral be \(I\). We can split it into two parts:
\[ I = \int_{-\pi/6}^{\pi/6} \frac{\pi}{1-\sin(|x|+\pi/6)} dx + \int_{-\pi/6}^{\pi/6} \frac{4x^{11}}{1-\sin(|x|+\pi/6)} dx = I_1 + I_2 \] Let's analyze \(I_2\). The integrand is \(f(x) = \frac{4x^{11}}{1-\sin(|x|+\pi/6)}\).
Let's check if it's even or odd: \(f(-x) = \frac{4(-x)^{11}}{1-\sin(|-x|+\pi/6)} = \frac{-4x^{11}}{1-\sin(|x|+\pi/6)} = -f(x)\).
Since the integrand is an odd function and the interval is symmetric, \(I_2 = 0\).
Now let's evaluate \(I_1\). The integrand is \(g(x) = \frac{\pi}{1-\sin(|x|+\pi/6)}\).
Let's check if it's even or odd: \(g(-x) = \frac{\pi}{1-\sin(|-x|+\pi/6)} = \frac{\pi}{1-\sin(|x|+\pi/6)} = g(x)\).
Since the integrand is an even function, we can write:
\[ I = I_1 = 2 \int_{0}^{\pi/6} \frac{\pi}{1-\sin(|x|+\pi/6)} dx \] For \(x \in [0, \pi/6]\), \(|x| = x\). So,
\[ I = 2\pi \int_{0}^{\pi/6} \frac{1}{1-\sin(x+\pi/6)} dx \] Let \(u = x + \pi/6\), so \(du = dx\). The limits of integration change:
When \(x=0\), \(u = \pi/6\).
When \(x=\pi/6\), \(u = \pi/3\).
\[ I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1}{1-\sin(u)} du \] Using the identity \(1-\sin(u) = 2\sin^2(\pi/4 - u/2)\):
\[ I = 2\pi \int_{\pi/6}^{\pi/3} \frac{1}{2\sin^2(\pi/4 - u/2)} du = \pi \int_{\pi/6}^{\pi/3} \csc^2(\pi/4 - u/2) du \] Now we integrate:
\[ \int \csc^2(\pi/4 - u/2) du = \frac{-\cot(\pi/4 - u/2)}{-1/2} = 2\cot(\pi/4 - u/2) \] Applying the limits:
\[ I = \pi \left[ 2\cot(\pi/4 - u/2) \right]_{\pi/6}^{\pi/3} \] \[ I = 2\pi \left[ \cot(\pi/4 - (\pi/3)/2) - \cot(\pi/4 - (\pi/6)/2) \right] \] \[ I = 2\pi \left[ \cot(\pi/4 - \pi/6) - \cot(\pi/4 - \pi/12) \right] \] \[ I = 2\pi \left[ \cot(\pi/12) - \cot(\pi/6) \right] \] We know \(\cot(\pi/6) = \sqrt{3}\) and \(\cot(\pi/12) = \cot(15^\circ) = 2+\sqrt{3}\).
\[ I = 2\pi \left[ (2+\sqrt{3}) - \sqrt{3} \right] = 2\pi(2) = 4\pi \] Step 4: Final Answer:
The value of the integral is \(4\pi\).