Mathematics - Indefinite Integration Question with Solution | TestHub

$I(x)=\int \frac{6dx}{{\sin }^{2}x(1-\cot x{)}^2}=\int \frac{6{\mathrm{cosec}}^{2}xdx}{(1-\cot x{)}^2}$
Put $1-\cot x=t$ ${\mathrm{cosec}}^2\mathrm{xdx}=\mathrm{dt}$ $\begin{array}{rl}& I=\int \frac{6dt}{t^2}=\frac{-6}{t}+c\\ & I(x)=\frac{-6}{1-\cot x}c,c=3\\ & I(x)=3-\frac{6}{1-\cot x},I\left(\frac{\pi }{12}\right)=3-\frac{6}{1-(2+\sqrt{3})}\\ & I\left(\frac{\pi }{12}\right)=3+\frac{6}{\sqrt{3}+1}=3+\frac{6(\sqrt{3}-1)}{2}=3\sqrt{3}\end{array}$