JEE Main 2026 Apr 6 Shift 1, Mathematics Q18: Some Properties of Definite Integrals

Call the integral $I$. Replace $x$ with $-x$ (the limits are symmetric) and add to the original. Since ${\cos }^{4}x$ is even and the denominators combine via $\frac{1}{1+e^{\sin x}}+\frac{1}{1+e^{-\sin x}}=1$:

$2I={\int }_{-\pi /4}^{\pi /4}32{\cos }^{4}xdx=2{\int }_0^{\pi /4}32{\cos }^{4}xdx$.

So $I=32{\int }_0^{\pi /4}{\cos }^{4}xdx$. Use ${\cos }^{4}x=\frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x$:

$I=32{\left[\frac{3x}{8}+\frac{\sin 2x}{4}+\frac{\sin 4x}{32}\right]}_0^{\pi /4}=32\left(\frac{3\pi }{32}+\frac{1}{4}+0\right)=3\pi +8$.