Definite IntegrationmediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Integral of sin^4 x + cos^4 x from 0 to 20π = 15π | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
The value of ∫020π(sin⁡4x+cos⁡4x) dx\displaystyle\int_0^{20\pi}(\sin^4x+\cos^4x)\,dx is equal to
A15π2\dfrac{15\pi}{2}
B25π25\pi
C15π15\picorrect
D25π2\dfrac{25\pi}{2}
Solution
Step 1: Simplify the integrand:
sin⁡4x+cos⁡4x=(sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x=1−2sin⁡2xcos⁡2x=1−sin⁡22x2.\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x=1-\frac{\sin^2 2x}{2}.
 Step 2: So
I=∫020π(1−sin⁡22x2)dx=20π−12∫020πsin⁡22x dx.I=\int_0^{20\pi}\left(1-\frac{\sin^2 2x}{2}\right)dx=20\pi-\frac12\int_0^{20\pi}\sin^2 2x\,dx.
 Step 3: The function sin⁡22x\sin^2 2x has period π2\dfrac{\pi}{2}, and 20π=40⋅π220\pi=40\cdot\dfrac{\pi}{2}, so
∫020πsin⁡22x dx=40∫0π/2sin⁡22x dx.\int_0^{20\pi}\sin^2 2x\,dx=40\int_0^{\pi/2}\sin^2 2x\,dx.
 Step 4: Evaluate ∫0π/2sin⁡22x dx=∫0π/21−cos⁡4x2 dx=[x2−sin⁡4x8]0π/2=π4.\displaystyle\int_0^{\pi/2}\sin^2 2x\,dx=\int_0^{\pi/2}\frac{1-\cos4x}{2}\,dx=\left[\frac{x}{2}-\frac{\sin4x}{8}\right]_0^{\pi/2}=\frac{\pi}{4}. Step 5: Therefore
I=20π−12⋅40⋅π4=20π−5π=15π.I=20\pi-\frac12\cdot40\cdot\frac{\pi}{4}=20\pi-5\pi=15\pi.
 Correct answer: (3)
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