JEE 2024 — Mathematics PYQ

\begin{aligned}
\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1 - \sin 2x} \, dx &= \alpha + \beta\sqrt{2} + \gamma\sqrt{3} \\
\text{let } \quad I &= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1 - \sin 2x} \, dx \\
&= \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} |\sin x - \cos x| \, dx \\
&= \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx \\
&= [\sin x + \cos x]_{\frac{\pi}{6}}^{\frac{\pi}{4}} - [\cos x + \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{3}}
\end{aligned}