The value of ∫_{-π/6}^{π/6} ( (π + 4x¹¹ ) / (1 - sin(|x| + π/6)) ) dx is equal to

The value of ∫_{-π/6}^{π/6} ( (π + 4x¹¹ ) / (1 - sin(|x| + π/6)) ) dx is equal to :

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

(A) \(8\pi\)

(B) \(4\pi\)

(C) \(2\pi\)

(D) \(6\pi\)

Correct Answer: (B) \(4\pi\)

Step-by-Step Solution

Step 1: Splitting the Integral

Given the integral, we can split the numerator into two separate 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 = I_1 + I_2

Step 2: Evaluating \(I_2\) (Odd Function Property)

Let's check the function \(f(x) = \frac{4x^{11}}{1 - \sin(|x| + \pi/6)}\). Since \(x^{11}\) is odd and \(|x|\) is even:

f(-x) = \frac{4(-x)^{11}}{1 - \sin(|-x| + \pi/6)} = -f(x)\) Since \(f(x)\) is an ODD function, the integral over symmetric limits is zero: \(I_2 = 0

Step 3: Evaluating \(I_1\) (Even Function Property)

The function \(g(x) = \frac{\pi}{1 - \sin(|x| + \pi/6)}\) is an EVEN function because \(g(-x) = g(x)\):

\(I_1 = 2 \int_{0}^{\pi/6} \frac{\pi}{1 - \sin(x + \pi/6)} dx\) (Using property: \(\int_{-a}^{a} even = 2\int_{0}^{a} even\))

Step 4: Using Trigonometric Identity

We use \(\sin \theta = \cos(\pi/2 - \theta)\) to simplify the denominator:

I_1 = 2\pi \int_{0}^{\pi/6} \frac{1}{1 - \cos(\pi/2 - (x + \pi/6))} dx\) \(I_1 = 2\pi \int_{0}^{\pi/6} \frac{1}{1 - \cos(\pi/3 - x)} dx\) Applying \(1 - \cos \theta = 2\sin^2(\theta/2)\): \(I_1 = 2\pi \int_{0}^{\pi/6} \frac{1}{2 \sin^2(\frac{\pi/3 - x}{2})} dx\) \(I_1 = \pi \int_{0}^{\pi/6} \csc^2\left(\frac{\pi}{6} - \frac{x}{2}\right) dx

Step 5: Integration and Limits

I_1 = \pi \left[ \frac{-\cot(\frac{\pi}{6} - \frac{x}{2})}{-1/2} \right]_{0}^{\pi/6}\) \(I_1 = 2\pi \left[ \cot\left(\frac{\pi}{6} - \frac{x}{2}\right) \right]_{0}^{\pi/6}\) \(I_1 = 2\pi \left[ \cot\left(\frac{\pi}{6} - \frac{\pi}{12}\right) - \cot\left(\frac{\pi}{6}\right) \right]\) \(I_1 = 2\pi [ \cot(15^\circ) - \cot(30^\circ) ]

Step 6: Final Calculation

Substitute values: \(\cot(15^\circ) = 2+\sqrt{3}\) and \(\cot(30^\circ) = \sqrt{3}\):

I = 2\pi [ (2 + \sqrt{3}) - \sqrt{3} ]\) \(I = 2\pi [ 2 ]\) Total Sum = \(4\pi

Related Theory

Definite Integral Properties: In JEE, the property of Even and Odd functions is crucial. For symmetric limits \([-a, a]\), an odd function results in \(0\), while an even function doubles over the interval \([0, a]\). This simplifies complex terms like \(x^{11}\) instantly.

Trigonometric Identities: Converting \(\sin\) to \(\cos\) to use the half-angle formula \(1 - \cos \theta = 2\sin^2(\theta/2)\) is a standard technique to handle integrands with \(1 \pm \sin x\) or \(1 \pm \cos x\) in the denominator.

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FAQs

Q1. Why is the second part of the integral zero? Because the integrand contains x¹¹ which is an odd function, and the limits are symmetric (-π/6 to π/6). Q2. What is the value of cot 15°? The value of cot 15° is 2 + √3.

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