JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 18

Symmetric limits. Adding the integral with $x\to-x$ removes the $e^{\sin x}$:
$$2I=\int_{-\pi/4}^{\pi/4}32\cos^4x\,dx,\qquad I=32\int_0^{\pi/4}\cos^4x\,dx.$$

Integrate. $\cos^4x=\tfrac18(3+4\cos2x+\cos4x)$, so
$$I=4\left[3x+2\sin2x+\tfrac14\sin4x\right]_0^{\pi/4}=3\pi+8.$$

(B)