JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 6

Remove $e^x$. For even $g$, $\displaystyle\int_{-a}^a\frac{g(x)}{1+e^x}dx=\int_0^ag(x)\,dx$ (add the integral to itself with $x\to-x$).

Simplify. $\sin^6x+\cos^6x=1-3\sin^2x\cos^2x=1-\frac34\sin^22x$, so the integral is $\displaystyle\int_0^{\pi/2}\frac{dx}{1-\frac34\sin^22x}=\int_0^{\pi/2}\frac{du}{1-\frac34\sin^2u}$ (with $u=2x$ and symmetry).

Standard result. $\displaystyle\int_0^{\pi/2}\frac{du}{1-k\sin^2u}=\frac\pi{2\sqrt{1-k}}=\frac\pi{2\cdot\frac12}=\pi$.

(C)