Question 283 marksShort AnswerSection C

If I1 = integral from -pi/4 to pi/4 of dx/(1 + cos 2x) and I2 = integral from -1/2 to 1/2 of |x| dx, then show that I1−4I2=0I_1 - 4 I_2 = 0.

Integrals
Definite Integrals — Properties and Evaluation

Official answer

Both integrals evaluate so that I1−4I2=0I_1 - 4I_2 = 0, as required.


I₁: Using 1+cos⁡2x=2cos⁡2x1+\cos 2x = 2\cos^2 x, I1=12∫−π/4π/4sec⁡2x dx=12[tan⁡x]−π/4π/4=12(1−(−1))=I_1 = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \sec^2x\, dx = \frac{1}{2}[\tan x]_{-\pi/4}^{\pi/4} = \frac{1}{2}(1-(-1)) = 1.


I₂: Since |x| is even, I2=2∫01/2x dx=2⋅[x2/2]01/2=2(1/8)=I_2 = 2\int_0^{1/2} x\, dx = 2\cdot[x^2/2]_0^{1/2} = 2(1/8) = 1/4.


Conclusion: I1−4I2=1−4(1/4)=1−1=I_1 - 4I_2 = 1 - 4(1/4) = 1 - 1 = 0.

Answer keywords:definite integraldouble angle formulaeven function propertysec squared integralmodulus function integral1+cos2x identity

Marking scheme

  1. 1.1 mark: correct simplification of I₁ using 1+cos⁡2x=2cos⁡2x1+\cos 2x = 2\cos^2 x and evaluating to 1.
  2. 2.1 mark: correct use of evenness of |x| to evaluate I2=1/4I_2 = 1/4.
  3. 3.1 mark: correct final computation I1−4I2=0I_1 - 4I_2 = 0 with conclusion.

Hint

Use 1+cos⁡2x=2cos⁡2x1+\cos 2x = 2\cos^2 x to simplify I₁ to ∫sec⁡2x dx\int \sec^2x\, dx; use evenness of |x| to write I2=2∫01/2x dxI_2 = 2\int_0^{1/2} x\, dx.

Quick oral answer

I₁ simplifies to 1 using sec⁡2x\sec^2x, and I₂ simplifies to 1/4 using the even-function property of |x|, so I1−4I2=0I_1 - 4I_2 = 0.

Common mistakes

  1. 1.Using 1−cos⁡2x=2sin⁡2x1-\cos 2x = 2\sin^2 x instead of the correct 1+cos⁡2x=2cos⁡2x1+\cos 2x = 2\cos^2 x, leading to a wrong antiderivative.
  2. 2.Integrating |x| across the full symmetric interval without splitting at x=0x=0, causing sign errors.
  3. 3.Arithmetic slip while multiplying 4×I24 \times I_2, leading to a nonzero (wrong) final result.

Analysis and explanation

This question combines a trigonometric simplification with the even-function property of definite integrals, both standard exam techniques.


Concept: For I₁, the identity 1+cos⁡2x=2cos⁡2x1+\cos 2x = 2\cos^2 x converts a non-standard integrand into sec⁡2x\sec^2x, whose antiderivative tan⁡x\tan x is elementary. For I₂, recognising |x| as even over the symmetric interval [−1/2,1/2][-1/2,1/2] allows halving the interval and dropping the modulus (since x≥0x \ge 0 on [0,1/2][0,1/2]).


Exam trap: A common mistake is integrating |x| directly over the full interval without splitting or using evenness, causing sign errors from the negative half. Another is misapplying the double-angle formula (using 1−cos⁡2x=2sin⁡2x1-\cos 2x=2\sin^2 x instead of 1+cos⁡2x=2cos⁡2x1+\cos 2x=2\cos^2 x).


Real-world link: Even/odd function properties of definite integrals are used extensively in Fourier analysis and signal processing to simplify integrals over symmetric time or frequency intervals.

Interesting facts

The property ∫−aaf(x) dx=2∫0af(x) dx\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx for even functions is one of the most frequently tested properties of definite integrals in CBSE board exams.

The double-angle identities used here trace back to the works of the 10th-century mathematician Abu al-Wafa, who compiled many trigonometric identities still used today.

Question details

Q28 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Integrals
Topic
Definite Integrals — Properties and Evaluation
Bloom's level
analyze
Syllabus topic
Unit 4: Integrals — Definite Integrals: Even Function Property and Trigonometric Simplification
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is 1+cos⁡2x1+\cos 2x replaced by 2cos⁡2x2\cos^2x?

This is the standard double-angle identity cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2x-1, rearranged to 1+cos⁡2x=2cos⁡2x1+\cos 2x = 2\cos^2 x, which converts the integrand into sec⁡2x\sec^2x, an easily integrable standard form.

Why can the modulus be dropped when evaluating I₂?

Because |x| is even, the integral over [−1/2,1/2][-1/2,1/2] equals twice the integral over [0,1/2][0,1/2], and on [0,1/2][0,1/2], x≥0x \ge 0 so ∣x∣=x|x| = x directly.