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 $I_1 - 4 I_2 = 0$ .

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

Both integrals evaluate so that $I_1 - 4I_2 = 0$, as required. **I₁:** Using $1+\cos 2x = 2\cos^2 x$, $I_1 = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \sec^2x\, dx = \frac{1}{2}[\tan x]_{-\pi/4}^{\pi/4} = \frac{1}{2}(1-(-1)) =$ **1**. **I₂:** Since |x| is even, $I_2 = 2\int_0^{1/2} x\, dx = 2\cdot[x^2/2]_0^{1/2} = 2(1/8) =$ **1/4**. **Conclusion:** $I_1 - 4I_2 = 1 - 4(1/4) = 1 - 1 =$ **0**.

Integrals

Definite Integrals — Properties and Evaluation

Official Answer

Both integrals evaluate so that $I_1 - 4I_2 = 0$ , as required.

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

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

Conclusion: $I_1 - 4I_2 = 1 - 4(1/4) = 1 - 1 =$ 0.

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

Marking Scheme

11 mark: correct simplification of I₁ using $1+\cos 2x = 2\cos^2 x$ and evaluating to 1.
21 mark: correct use of evenness of |x| to evaluate $I_2 = 1/4$ .
31 mark: correct final computation $I_1 - 4I_2 = 0$ with conclusion.

Hint

Use $1+\cos 2x = 2\cos^2 x$ to simplify I₁ to $\int \sec^2x\, dx$ ; use evenness of |x| to write $I_2 = 2\int_0^{1/2} x\, dx$ .

Quick Oral Answer

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

Analysis & Explanation

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

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

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 $1-\cos 2x=2\sin^2 x$ instead of $1+\cos 2x=2\cos^2 x$ ).

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.

Common Mistakes

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

Interesting Facts

The property $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$ 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.

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Frequently Asked Questions

Why is $1+\cos 2x$ replaced by $2\cos^2x$ ?

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

Why can the modulus be dropped when evaluating I₂?

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