Question #be320

2 Answers
Jun 22, 2017

-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

Jun 22, 2017

-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.