JEE Main 2024 (Online) 8th April Morning ShiftQuestion 1
Let $I(x)=\int\frac{6}{\sin^{2}x(1-\cot x)^{2}}dx$. If $I(0)=3$, then $I\left(\frac{\pi}{12}\right)$ is equal to
Indefinite Integration
Solution
Concept. $\frac{dx}{\sin^2x}=d(1-\cot x)$, so the integral is $\int\frac{6\,du}{u^2}$. Formulas used. (1) $u=1-\cot x$: $du=\operatorname{cosec}^2x\,dx$(2) $\cot\frac\pi{12}=2+\sqrt3$Step 1 - Integrate. By (1), $$I(x)=\int\frac{6\,du}{u^2}=-\frac{6}{1-\cot x}+C.$$ Step 2 - Constant. As $x\to0^+$, $\cot x\to\infty$ and $-\frac{6}{1-\cot x}\to0$. So $C=I(0)=3$. Step 3 - Value. By (2), $1-\cot\frac\pi{12}=-(1+\sqrt3)$: $$I\Big(\frac\pi{12}\Big)=\frac{6}{1+\sqrt3}+3=3(\sqrt3-1)+3=3\sqrt3.$$ Answer: (B)