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MCQMediumJEE 2026Properties of Definite Integrals

JEE Mathematics 2026 Question with Solution

The value of ∫−π/6π/6π+4x111−sin⁡(∣x∣+π/6) dx\int_{-\pi/6}^{\pi/6} \frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)} \, dx is equal to:

  • A

    8π8\pi

  • B

    2π2\pi

  • C

    6π6\pi

  • D

    4π4\pi

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Evaluate

I=∫−π/6π/6π+4x111−sin⁡(∣x∣+π/6) dxI=\int_{-\pi/6}^{\pi/6} \frac{\pi+4x^{11}}{1-\sin(|x|+\pi/6)} \, dx

Find: The value of the definite integral and the correct option.

Split the integral into two parts:

I=∫−π/6π/6π1−sin⁡(∣x∣+π/6) dx+∫−π/6π/64x111−sin⁡(∣x∣+π/6) dxI=\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

Let these be I1I_1 and I2I_2 respectively.

For

I2=∫−π/6π/64x111−sin⁡(∣x∣+π/6) dxI_2=\int_{-\pi/6}^{\pi/6} \frac{4x^{11}}{1-\sin(|x|+\pi/6)} \, dx

the denominator depends on ∣x∣|x|, so it is even, while x11x^{11} is odd. Hence the whole integrand is odd:

f(−x)=4(−x)111−sin⁡(∣−x∣+π/6)=−4x111−sin⁡(∣x∣+π/6)=−f(x)f(-x)=\frac{4(-x)^{11}}{1-\sin(|-x|+\pi/6)}=-\frac{4x^{11}}{1-\sin(|x|+\pi/6)}=-f(x)

Therefore, over the symmetric interval [−π/6,π/6][-\pi/6,\pi/6],

I2=0I_2=0

Now consider

I1=∫−π/6π/6π1−sin⁡(∣x∣+π/6) dxI_1=\int_{-\pi/6}^{\pi/6} \frac{\pi}{1-\sin(|x|+\pi/6)} \, dx

This integrand is even, so

I=I1=2∫0π/6π1−sin⁡(x+π/6) dxI=I_1=2\int_0^{\pi/6} \frac{\pi}{1-\sin(x+\pi/6)} \, dx

Hence,

I=2π∫0π/611−sin⁡(x+π/6) dxI=2\pi\int_0^{\pi/6} \frac{1}{1-\sin(x+\pi/6)} \, dx

Put

u=x+π/6u=x+\pi/6

so that

du=dxdu=dx

The limits change as:

  • when x=0x=0, u=π/6u=\pi/6
  • when x=π/6x=\pi/6, u=π/3u=\pi/3

Thus,

I=2π∫π/6π/311−sin⁡u duI=2\pi\int_{\pi/6}^{\pi/3} \frac{1}{1-\sin u} \, du

Using

1−sin⁡u=2sin⁡2(π/4−u/2)1-\sin u=2\sin^2\left(\pi/4-u/2\right)

we get

I=2π∫π/6π/312sin⁡2(π/4−u/2) duI=2\pi\int_{\pi/6}^{\pi/3} \frac{1}{2\sin^2(\pi/4-u/2)} \, du

that is,

I=π∫π/6π/3csc⁡2(π/4−u/2) duI=\pi\int_{\pi/6}^{\pi/3} \csc^2\left(\pi/4-u/2\right) \, du

Now,

∫csc⁡2(π/4−u/2) du=2cot⁡(π/4−u/2)\int \csc^2\left(\pi/4-u/2\right) \, du=2\cot\left(\pi/4-u/2\right)

Therefore,

I=π[2cot⁡(π/4−u/2)]π/6π/3I=\pi\left[2\cot\left(\pi/4-u/2\right)\right]_{\pi/6}^{\pi/3}

So,

I=2π[cot⁡(π/4−π/6)−cot⁡(π/4−π/12)]I=2\pi\left[\cot\left(\pi/4-\pi/6\right)-\cot\left(\pi/4-\pi/12\right)\right] I=2π[cot⁡(π/12)−cot⁡(π/6)]I=2\pi\left[\cot(\pi/12)-\cot(\pi/6)\right]

Using

cot⁡(π/6)=3,cot⁡(π/12)=2+3\cot(\pi/6)=\sqrt{3}, \qquad \cot(\pi/12)=2+\sqrt{3}

we obtain

I=2π[(2+3)−3]=2π⋅2=4πI=2\pi\left[(2+\sqrt{3})-\sqrt{3}\right]=2\pi\cdot 2=4\pi

Therefore, the value of the integral is 4π4\pi and the correct option is D.

Common mistakes

  • Treating the entire integrand as odd because of the term x11x^{11} is incorrect. The numerator is split as π+4x11\pi+4x^{11}, so only the second part gives an odd integrand. First separate the integral into even and odd parts.

  • Ignoring the effect of ∣x∣|x| in the denominator leads to a wrong parity check. Since ∣−x∣=∣x∣|{-x}|=|x|, the denominator is even. Use this carefully before applying the symmetric-interval properties.

  • While substituting u=x+π/6u=x+\pi/6, students often forget to change the limits from 00 and π/6\pi/6 to π/6\pi/6 and π/3\pi/3. In a definite integral, always update the limits after substitution instead of reverting back later.

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