## Thomas' Calculus 13th Edition

$-\dfrac{1}{2} \cos 2x +\cot x+C$
Solve. $\int (\sin 2x ) dx - \int (\csc^2 x) dx$ Thus, we have $\int (\sin 2x ) dx - \int (\csc^2 x) dx=-\dfrac{1}{2} \cos 2x -(-\cot x)+C$ or, $=-\dfrac{1}{2} \cos 2x +\cot x+C$