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

Split: $\int_{-\pi/6}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx + \int_{-\pi/6}^{\pi/6} \frac{4x^{11}}{1 - \sin(|x| + \pi/6)} dx$. Second integral = 0 (odd function over symmetric interval). First integral: $2\pi \int_{0}^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$. Let $u = x + \pi/6$: $= 2\pi \int_{\pi/6}^{\pi/3} \frac{1}{1 - \sin u} du$. $\frac{1}{1-\sin u} = \sec^2 u + \sec u \tan u$. $= 2\pi [\tan u + \sec u]_{\pi/6}^{\pi/3} = 2\pi[(\sqrt{3}+2) - \sqrt{3}] = 4\pi$.

JEE Main 2026 — Mathematics Calculus

2026mcqmedium

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

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

A
$8 \pi$
B
$6 \pi$
C
$2 \pi$
D
$4 \pi$

Did you get this right?

Solution

Split: $\displaystyle \int_{-\pi/6}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx + \int_{-\pi/6}^{\pi/6} \frac{4x^{11}}{1 - \sin(|x| + \pi/6)} dx$ .

Second integral = 0 (odd function over symmetric interval).

First integral: $\displaystyle 2\pi \int_{0}^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$ .

Let $u = x + \pi/6$ : $\displaystyle = 2\pi \int_{\pi/6}^{\pi/3} \frac{1}{1 - \sin u} du$ .

$\frac{1}{1-\sin u} = \sec^2 u + \sec u \tan u$ .

$= 2\pi [\tan u + \sec u]_{\pi/6}^{\pi/3} = 2\pi[(\sqrt{3}+2) - \sqrt{3}] = 4\pi$ .

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JEE Main 2026 — Mathematics Calculus

2026mcqmedium

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

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

A
$8 \pi$
B
$6 \pi$
C
$2 \pi$
D
$4 \pi$

Did you get this right?

Solution

Split: $\displaystyle \int_{-\pi/6}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx + \int_{-\pi/6}^{\pi/6} \frac{4x^{11}}{1 - \sin(|x| + \pi/6)} dx$ .

Second integral = 0 (odd function over symmetric interval).

First integral: $\displaystyle 2\pi \int_{0}^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx$ .

Let $u = x + \pi/6$ : $\displaystyle = 2\pi \int_{\pi/6}^{\pi/3} \frac{1}{1 - \sin u} du$ .

$\frac{1}{1-\sin u} = \sec^2 u + \sec u \tan u$ .

$= 2\pi [\tan u + \sec u]_{\pi/6}^{\pi/3} = 2\pi[(\sqrt{3}+2) - \sqrt{3}] = 4\pi$ .

Your note

JEE Main Calculus in other years

Calculus 2025 Calculus 2024 Calculus 2023 Calculus 2022 Calculus 2021 Calculus 2020 Calculus 2019 Calculus 2018 Calculus 2017 Calculus 2016 Calculus 2015 Calculus 2014

Keep practicing Calculus

Work through every JEE Main Calculus PYQ, year by year.

Browse all Mathematics topics

JEE Main 2026 — Mathematics Calculus

Options

Did you get this right?

Solution

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JEE Main Calculus in other years

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JEE Main 2026 — Mathematics Calculus

Options

Did you get this right?

Solution

Your note

JEE Main Calculus in other years

More JEE Main Calculus questions

Other Mathematics topics for JEE Main

Algebra

Coordinate Geometry

Vectors & 3D Geometry

Probability & Statistics

Trigonometry

Keep practicing Calculus