Why does the term vanish?

Because satisfies , making it an odd function integrated over a symmetric interval .

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Integral Calculus: JEE Main Mathematics Question with Solution

The value of the integral

\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x

is :

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\frac{\pi^2}{12 \sqrt{3}}

\frac{\pi^2}{6 \sqrt{3}}

\frac{\pi^2}{6}

\frac{\pi^2}{3 \sqrt{3}}

Step-by-step solutionView

Correct answer

Splitting the integral into odd and even components and applying the substitution $t = an x$ evaluates the integral to $\frac{\pi^2}{6\sqrt{3}}$ .

Option analysis

Why each option works or fails

A ·

\frac{\pi^2}{12 \sqrt{3}}

Believing that an extra factor of $1/2$ is introduced when using symmetry properties of even functions on symmetric limits. Remember that $\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$ for an even function, not $\int_0^a f(x)\,dx$ .

B ·

\frac{\pi^2}{6 \sqrt{3}}

None. This is the correct value obtained by using odd/even function symmetry and standard tangent half-angle/substitution methods. Correctly split the integral: the odd part vanishes, and the remaining integral reduces to $\frac{\pi}{2} \int_0^{\pi/4} \frac{dx}{2-\cos 2x} = \frac{\pi^2}{6\sqrt{3}}$ .

C ·

\frac{\pi^2}{6}

Forgetting the $\sqrt{3}$ factor that emerges from evaluating $\frac{1}{\sqrt{a^2}} \arctan(u/a)$ when integrating forms like $\frac{1}{u^2 + 3}$ . Recall the formula $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$ , where $a = \sqrt{3}$ here.

D ·

\frac{\pi^2}{3 \sqrt{3}}

Omitting the factor of $1/2$ when transforming $dx$ to $dt$ or during the standard integral $\int \frac{dt}{1 + 3t^2}$ . Carefully trace differential substitutions: writing $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$ and substituting $t = \tan x$ requires $dx = \frac{dt}{1+t^2}$ .

Reviewed route

Solution

StepWorking

01given

Integral $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} dx$ .

02goal

Evaluate the value of the definite integral.

03approach

Split the integrand into $x/(2-\cos 2x)$ and $(\pi/4)/(2-\cos 2x)$ . The first is an odd function over $[-\pi/4, \pi/4]$ so it vanishes. The second is an even function, which can be simplified using $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$ and the substitution $t = \tan x$ .

04execute

Using symmetry: $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x}{2 - \cos 2x} dx + \frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} dx = 0 + 2 \cdot \frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} dx = \frac{\pi}{2} \int_0^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} dx$ Express $\cos 2x$ in terms of $\tan x$ : $2 - \cos 2x = 2 - \frac{1 - \tan^2 x}{1 + \tan^2 x} = \frac{2(1+\tan^2 x) - (1-\tan^2 x)}{1+\tan^2 x} = \frac{1 + 3\tan^2 x}{\sec^2 x}$ Thus, $I = \frac{\pi}{2} \int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{1 + 3\tan^2 x} dx$ Substitute $t = \tan x$ , $dt = \sec^2 x dx$ , with limits $t \in [0, 1]$ : $I = \frac{\pi}{2} \int_0^1 \frac{dt}{1 + 3t^2} = \frac{\pi}{2} \left[ \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}t) \right]_0^1 = \frac{\pi}{2\sqrt{3}} \tan^{-1}(\sqrt{3}) = \frac{\pi}{2\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi^2}{6\sqrt{3}}$

✓verify

The integrand $\frac{x+\pi/4}{2-\cos 2x} > 0$ on $(-\pi/4, \pi/4)$ , and width is $\pi/2$ , average value is around $1/2$ , yielding an integral around $\pi/4 \approx 0.8$ , consistent with $\pi^2/(6\sqrt{3}) \approx 0.95$ .

Hints that build this answer step by step

How can the given integral $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} \, dx$ be simplified using symmetry over the symmetric interval $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$ ?
Split the numerator: $\frac{x}{2-\cos 2x}$ is an odd function (integrates to $0$ ), leaving $I = 2 \cdot \frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{dx}{2-\cos 2x} = \frac{\pi}{2} \int_0^{\frac{\pi}{4}} \frac{dx}{2-\cos 2x}$ .
Expressing $\cos 2x$ in terms of $t = \tan x$ , what does the integral $J = \int_0^{\frac{\pi}{4}} \frac{dx}{2-\cos 2x}$ transform into?
$J = \int_0^1 \frac{dt}{1 + 3t^2}$
What is the value of $\int_0^1 \frac{dt}{1+3t^2}$ , and what is the final value of $I = \frac{\pi}{2} J$ ?
$J = \frac{\pi}{3\sqrt{3}}$ , so $I = \frac{\pi}{2} \cdot \frac{\pi}{3\sqrt{3}} = \frac{\pi^2}{6\sqrt{3}}$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why does the

\int_{-\pi/4}^{\pi/4} \frac{x}{2-\cos 2x} dx

term vanish?

Because $f(x) = \frac{x}{2-\cos 2x}$ satisfies $f(-x) = -f(x)$ , making it an odd function integrated over a symmetric interval $[-a, a]$ .

Answer

Splitting the integral into odd and even components and applying the substitution $t = an x$ evaluates the integral to $\frac{\pi^2}{6\sqrt{3}}$ .

Why each option works or fails

A: $\frac{\pi^2}{12 \sqrt{3}}$ - Believing that an extra factor of $1/2$ is introduced when using symmetry properties of even functions on symmetric limits. Remember that $\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$ for an even function, not $\int_0^a f(x)\,dx$ .
B · correct: $\frac{\pi^2}{6 \sqrt{3}}$ - None. This is the correct value obtained by using odd/even function symmetry and standard tangent half-angle/substitution methods. Correctly split the integral: the odd part vanishes, and the remaining integral reduces to $\frac{\pi}{2} \int_0^{\pi/4} \frac{dx}{2-\cos 2x} = \frac{\pi^2}{6\sqrt{3}}$ .
C: $\frac{\pi^2}{6}$ - Forgetting the $\sqrt{3}$ factor that emerges from evaluating $\frac{1}{\sqrt{a^2}} \arctan(u/a)$ when integrating forms like $\frac{1}{u^2 + 3}$ . Recall the formula $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$ , where $a = \sqrt{3}$ here.
D: $\frac{\pi^2}{3 \sqrt{3}}$ - Omitting the factor of $1/2$ when transforming $dx$ to $dt$ or during the standard integral $\int \frac{dt}{1 + 3t^2}$ . Carefully trace differential substitutions: writing $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$ and substituting $t = \tan x$ requires $dx = \frac{dt}{1+t^2}$ .

Step-by-step solution

given: Integral $\frac{\pi^2}{6\sqrt{3}}$ .
goal: Evaluate the value of the definite integral.
approach: Split the integrand into $x/(2-\cos 2x)$ and $(\pi/4)/(2-\cos 2x)$ . The first is an odd function over $[-\pi/4, \pi/4]$ so it vanishes. The second is an even function, which can be simplified using $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$ and the substitution $t = \tan x$ .
execute: Using symmetry: $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x}{2 - \cos 2x} dx + \frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} dx = 0 + 2 \cdot \frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} dx = \frac{\pi}{2} \int_0^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} dx$ Express $\cos 2x$ in terms of $\tan x$ : $2 - \cos 2x = 2 - \frac{1 - \tan^2 x}{1 + \tan^2 x} = \frac{2(1+\tan^2 x) - (1-\tan^2 x)}{1+\tan^2 x} = \frac{1 + 3\tan^2 x}{\sec^2 x}$ Thus, $I = \frac{\pi}{2} \int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{1 + 3\tan^2 x} dx$ Substitute $t = \tan x$ , $dt = \sec^2 x dx$ , with limits $t \in [0, 1]$ : $I = \frac{\pi}{2} \int_0^1 \frac{dt}{1 + 3t^2} = \frac{\pi}{2} \left[ \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}t) \right]_0^1 = \frac{\pi}{2\sqrt{3}} \tan^{-1}(\sqrt{3}) = \frac{\pi}{2\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi^2}{6\sqrt{3}}$
verify: The integrand $\frac{x+\pi/4}{2-\cos 2x} > 0$ on $(-\pi/4, \pi/4)$ , and width is $\pi/2$ , average value is around $1/2$ , yielding an integral around $\pi/4 \approx 0.8$ , consistent with $\pi^2/(6\sqrt{3}) \approx 0.95$ .

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

Splitting the integral into odd and even components and applying the substitution $t = an x$ evaluates the integral to $\frac{\pi^2}{6\sqrt{3}}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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