Question

$\int \frac{\cos 2x}{\sin x}dx$

Answer 1

$I=\int \frac{\cos 2x}{\sin x}dx\because \cos 2x={\cos }^{2}x-{\sin }^{2}x=1-{\sin }^{2}x-{\sin }^{2}x=1-2{\sin }^{2}x\therefore I=\int \frac{1-2{\sin }^{2}x}{\sin x}dx=\int [\frac{1}{\sin x}-\frac{2{\sin }^{2}x}{\sin x}]dx⟹=\int \csc xdx-2\int \sin xdx⟹=-\ln |\csc x+\cot x|-2(-\cos x)+CI=2\cos x-\ln |\csc x+\cot x|+C$