Test Index

Definite Integrals Part 4

Section: Mathematics

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Question : 30 of 31

Marks: +1, -0

The value of

\int\limits_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \left( \;\frac{\pi+4 x^{11}}{1-\sin\left(|x|+\frac{\pi}{6}\right)} \right) dx

[21 Jan 2026 Shift 1]

$8\pi$
$4\pi$
$2\pi$
$6\pi$

Solution:

Concept:

Use the property

\int_{-a}^{a} f(x)dx = \int_{0}^{a} [f(x)+f(-x)]dx

.

Since

|x|

is even, only the numerator changes sign for

x^{11}

.

Explanation:

Let

I = \int_{-\pi/6}^{\pi/6} \frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)}dx

.

Write

I = \int_{0}^{\pi/6} \frac{\pi+4x^{11}}{1-\sin(x+\pi/6)}dx + \int_{0}^{\pi/6} \frac{\pi-4x^{11}}{1-\sin(x+\pi/6)}dx

(since

|{-x}| = |x|

and

(-x)^{11} = -x^{11}

).

The sum simplifies to

I = 2\pi \int_{0}^{\pi/6} \frac{dx}{1-\sin(x+\pi/6)}

.

Rationalise: multiply numerator and denominator by

1+\sin(x+\pi/6)

:

\frac{1}{1-\sin\theta} = \frac{1+\sin\theta}{1-\sin^2\theta} = \frac{1+\sin\theta}{\cos^2\theta} = \sec^2\theta + \sec\theta \tan\theta

where

\theta = x+\pi/6

.

Integrate:

\int (\sec^2\theta + \sec\theta \tan\theta) d\theta = \tan\theta + \sec\theta

.

Thus

I = 2\pi [\tan(x+\pi/6) + \sec(x+\pi/6)]_{0}^{\pi/6}

.

At upper limit

x=\pi/6

:

\tan(\pi/3)+\sec(\pi/3)=\sqrt{3}+2

.

At lower limit

x=0

:

\tan(\pi/6)+\sec(\pi/6)=\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}=\sqrt{3}

.

Difference =

(\sqrt{3}+2) - \sqrt{3} = 2

.

So

I = 2\pi \times 2 = 4\pi

.

Answer:

4\pi

, which corresponds to option B.

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