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Answer 1 · 2017-06-22T17:22:31

#-4cot2x+3cotx+C#

#int(2-6cos2x)/(sin^2 2x)dx#

#=int(2-6(2cos^2x-1))/(2sinxcosx)^2dx#

#=int(8-12cos^2x)/(4sin^2xcos^2x)dx#

#=int8/(sin^2 2x)dx-3int1/sin^2xdx#

#=8intcsc^2 2xdx-3intcsc^2xdx#

Keep in mind a standard derivative of a trigonometric function: #d/dxcotx=-csc^2x#
Therefore the antiderivative of #csc^2x=-cotx#

#8intcsc^2 2xdx-3intcsc^2xdx#

#=8intcsc^2 2x*(d(2x))/2-3intcsc^2xdx#

#=-4cot2x+3cotx+C#

Answer 2 · 2017-06-22T18:04:32

# -cot(2x)+3csc(2x)+C.#

#int(2-6cos2x)/sin^2(2x)dx=int{2/sin^2(2x)-6(cos(2x)/sin(2x))(1/sin(2x))}dx,#

#=int{2csc^2(2x)-6cot(2x)csc(2x)}dx,#

#=-2cot(2x)/2-6(-csc(2x))/2,#

#=-cot(2x)+3csc(2x)+C.#