JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Definite Integrals

Separate the integrand as $\frac{2}{\sin x(1+\cos x)}+\frac{3}{1+\cos x}$. Using $\frac{1}{1+\cos x}=\frac{1}{2}{\sec }^2\frac{x}{2}$, the second part integrates to $3\tan \frac{x}{2}$, which from $\frac{\pi }{3}$ to $\frac{\pi }{2}$ gives $3\left(1-\frac{1}{\sqrt{3}}\right)=3-\sqrt{3}$. For the first part put $t=\tan \frac{x}{2}$, so $\frac{2dx}{\sin x(1+\cos x)}=\left(\frac{1}{t}+t\right)dt$, which integrates to $\ln t+\frac{t^2}{2}$; between $t=\frac{1}{\sqrt{3}}$ and $t=1$ this equals $\frac{1}{3}+\ln \sqrt{3}$. Combining the two parts, the integral equals $\frac{10}{3}-\sqrt{3}+\ln \sqrt{3}$.