JEE Main 2026 — Mathematics Calculus

Let $\displaystyle I = \int_0^{20\pi} (\sin^4 x + \cos^4 x) \, dx$I=∫020π​(sin4x+cos4x)dx

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

$$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$$sin4x+cos4x=(sin2x+cos2x)2−2sin2xcos2x

Since $\sin^2 x + \cos^2 x = 1$sin2x+cos2x=1 and $2\sin x \cos x = \sin 2x$2sinxcosx=sin2x:

$$\sin^4 x + \cos^4 x = 1 - \dfrac{1}{2} (2\sin x \cos x)^2 = 1 - \dfrac{1}{2} \sin^2 2x$$sin4x+cos4x=1−21​(2sinxcosx)2=1−21​sin22x

Using the half-angle formula $\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}$sin2θ=21−cos2θ​:

$$\sin^4 x + \cos^4 x = 1 - \dfrac{1}{2} \left( \dfrac{1 - \cos 4x}{2} \right) = 1 - \dfrac{1}{4} + \dfrac{1}{4} \cos 4x = \dfrac{3}{4} + \dfrac{1}{4} \cos 4x$$sin4x+cos4x=1−21​(21−cos4x​)=1−41​+41​cos4x=43​+41​cos4x

Substituting this back into the integral:

$$I = \int_0^{20\pi} \left( \dfrac{3}{4} + \dfrac{1}{4} \cos 4x \right) \, dx$$I=∫020π​(43​+41​cos4x)dx

$$I = \left[ \dfrac{3}{4} x + \dfrac{1}{16} \sin 4x \right]_0^{20\pi}$$I=[43​x+161​sin4x]020π​

Evaluating the limits:

$$I = \left( \dfrac{3}{4} (20\pi) + \dfrac{1}{16} \sin(80\pi) \right) - (0 + 0)$$I=(43​(20π)+161​sin(80π))−(0+0)

Since $\sin(80\pi) = 0$sin(80π)=0:

$$I = 15\pi$$I=15π

Answer: $15\pi$15π