JEE Main 2026 — Mathematics Calculus

Let $\displaystyle I = \int_{-\pi/4}^{\pi/4} \dfrac{32\cos^4 x}{1 + e^{\sin x}} dx$I=∫−π/4π/4​1+esinx32cos4x​dx

Using the definite integral property $\displaystyle \int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$∫−aa​f(x)dx=∫0a​(f(x)+f(−x))dx, we get:

$$I = \int_{0}^{\pi/4} \left( \dfrac{32\cos^4 x}{1 + e^{\sin x}} + \dfrac{32\cos^4(-x)}{1 + e^{\sin(-x)}} \right) dx$$I=∫0π/4​(1+esinx32cos4x​+1+esin(−x)32cos4(−x)​)dx

$$I = \int_{0}^{\pi/4} \left( \dfrac{32\cos^4 x}{1 + e^{\sin x}} + \dfrac{32\cos^4 x}{1 + e^{-\sin x}} \right) dx$$I=∫0π/4​(1+esinx32cos4x​+1+e−sinx32cos4x​)dx

$$I = \int_{0}^{\pi/4} \left( \dfrac{32\cos^4 x}{1 + e^{\sin x}} + \dfrac{32\cos^4 x \cdot e^{\sin x}}{e^{\sin x} + 1} \right) dx$$I=∫0π/4​(1+esinx32cos4x​+esinx+132cos4x⋅esinx​)dx

$$I = \int_{0}^{\pi/4} 32\cos^4 x \left( \dfrac{1 + e^{\sin x}}{1 + e^{\sin x}} \right) dx$$I=∫0π/4​32cos4x(1+esinx1+esinx​)dx

$$I = \int_{0}^{\pi/4} 32\cos^4 x dx$$I=∫0π/4​32cos4xdx

Using the trigonometric identity $\cos^2 x = \dfrac{1 + \cos 2x}{2}$cos2x=21+cos2x​, we can write:

$$\cos^4 x = \left( \dfrac{1 + \cos 2x}{2} \right)^2 = \dfrac{1}{4} (1 + 2\cos 2x + \cos^2 2x)$$cos4x=(21+cos2x​)2=41​(1+2cos2x+cos22x)

$$\cos^4 x = \dfrac{1}{4} \left( 1 + 2\cos 2x + \dfrac{1 + \cos 4x}{2} \right) = \dfrac{3}{8} + \dfrac{1}{2}\cos 2x + \dfrac{1}{8}\cos 4x$$cos4x=41​(1+2cos2x+21+cos4x​)=83​+21​cos2x+81​cos4x

Substituting this back into the integral:

$$I = 32 \int_{0}^{\pi/4} \left( \dfrac{3}{8} + \dfrac{1}{2}\cos 2x + \dfrac{1}{8}\cos 4x \right) dx$$I=32∫0π/4​(83​+21​cos2x+81​cos4x)dx

$$I = \int_{0}^{\pi/4} (12 + 16\cos 2x + 4\cos 4x) dx$$I=∫0π/4​(12+16cos2x+4cos4x)dx

Integrating term by term:

$$I = \left[ 12x + 8\sin 2x + \sin 4x \right]_{0}^{\pi/4}$$I=[12x+8sin2x+sin4x]0π/4​

$$I = \left( 12\left(\dfrac{\pi}{4}\right) + 8\sin\left(\dfrac{\pi}{2}\right) + \sin(\pi) \right) - (0 + 0 + 0)$$I=(12(4π​)+8sin(2π​)+sin(π))−(0+0+0)

$$I = 3\pi + 8(1) + 0 = 3\pi + 8$$I=3π+8(1)+0=3π+8

Answer: $3\pi + 8$3π+8