Question

Question #be320

Answer 1

`-4cot2x+3cotx+C`

Explanation:

`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

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

Explanation:

`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.`