The value of int 0 20π (sin 4 x + cos 4 x) d x is equal to:

Mathematics · Definite Integration

JEE Main 2026 — 2 April, Evening Shift — Question 39

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

Option A:
$\frac{15\pi}{2}$
Option B:
$25\pi$
Option C:
$15\pi$
Correct
Option D:
$\frac{25\pi}{2}$

Answer: C

Step-by-step solution

$I=\int_{0}^{20 \pi}\left(1-2 \sin ^{2} x \cos ^{2} x\right) \cdot d x$

=20 \pi-\frac{1}{2} \int_{0}^{20 \pi} \sin ^{2} 2 x d x

$20 \pi-\frac{1}{2}(40) \int_{0}^{\frac{\pi}{2}} \sin ^{2} 2 \mathrm{x}$ $20 \pi-\frac{20}{2}\left(x-\frac{\sin 2 x}{2}\right)_{0}^{\frac{\pi}{2}}$ $20 \pi-10\left(\frac{\pi}{2}\right)=15 \pi$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: Definite Integration
Topic: Methods of solving definite integrals(kings rule,odd even)