JEE Main 2026 Apr 2 Shift 2, Mathematics Q17: Some Properties of Definite Integrals

JEE Main2026Apr 2, Shift 2Mathematics

The value of $\int_{0}^{20 π} (sin^{4} x + cos^{4} x) d x$ is equal to:

Solution

Use the identity $sin^{4} x + cos^{4} x = 1 - 2 sin^{2} x cos^{2} x = 1 - \frac{sin ^{2} 2 x}{2}$ .

Then $I = \int_{0}^{20 π} (1 - \frac{sin ^{2} 2 x}{2}) d x = 20 π - \frac{1}{2} \int_{0}^{20 π} sin^{2} 2 x d x$ .

By periodicity, $\int_{0}^{20 π} sin^{2} 2 x d x = 40 \cdot \int_{0}^{π /2} sin^{2} 2 x d x = 40 \cdot \frac{π}{4} = 10 π$ .

$∴ I = 20 π - \frac{1}{2} (10 π) = 15 π$ .

Concept behind this question

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