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JEE Main 20262 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2026 · 2 April · Evening Shift · Mathematics · Definite Integration

The value of ∫_0^ 20π ( ^4 x + ^4 x) \, dx is equal to:

The value of ∫020π(sin⁡4x+cos⁡4x) dx\int_0^{20\pi} (\sin^4 x + \cos^4 x) \, dx is equal to:

Options

  1. A15π2\dfrac{15\pi}{2}
  2. B25π25\pi
  3. C15π15\pi
  4. D25π2\dfrac{25\pi}{2}

Correct answer

C. 15π15\pi

Step-by-step solution

Let I=∫020π(sin⁡4x+cos⁡4x) dxI = \int_0^{20\pi} (\sin^4 x + \cos^4 x) \, dx

Using the algebraic identity a2+b2=(a+b)2−2aba^2 + b^2 = (a+b)^2 - 2ab, the integrand can be simplified:

sin⁡4x+cos⁡4x=(sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x

Since sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 and 2sin⁡xcos⁡x=sin⁡2x2\sin x \cos x = \sin 2x:

sin⁡4x+cos⁡4x=1−12(2sin⁡xcos⁡x)2=1−12sin⁡22x\sin^4 x + \cos^4 x = 1 - \dfrac{1}{2} (2\sin x \cos x)^2 = 1 - \dfrac{1}{2} \sin^2 2x

Using the half-angle formula sin⁡2θ=1−cos⁡2θ2\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}:

sin⁡4x+cos⁡4x=1−12(1−cos⁡4x2)=1−14+14cos⁡4x=34+14cos⁡4x\sin^4 x + \cos^4 x = 1 - \dfrac{1}{2} \left( \dfrac{1 - \cos 4x}{2} \right) = 1 - \dfrac{1}{4} + \dfrac{1}{4} \cos 4x = \dfrac{3}{4} + \dfrac{1}{4} \cos 4x

Substituting this back into the integral:

I=∫020π(34+14cos⁡4x) dxI = \int_0^{20\pi} \left( \dfrac{3}{4} + \dfrac{1}{4} \cos 4x \right) \, dx

I=[34x+116sin⁡4x]020πI = \left[ \dfrac{3}{4} x + \dfrac{1}{16} \sin 4x \right]_0^{20\pi}

Evaluating the limits:

I=(34(20π)+116sin⁡(80π))−(0+0)I = \left( \dfrac{3}{4} (20\pi) + \dfrac{1}{16} \sin(80\pi) \right) - (0 + 0)

Since sin⁡(80π)=0\sin(80\pi) = 0:

I=15πI = 15\pi

Answer: 15π15\pi
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