Question

what is the antiderivate of 4sin^2 x - (csc^2x/4)

Answer 1

\[4\sin{^2}x - \frac{ \csc{^2}x }{ 4 }\]

Answer 2

sin^2 = 1 - cos^2
cos(2x) = cos^2 - sin^2 = cos^2 - ( 1 - cos^2)
cos(2x) = 2cos^2 - 1
cos^2 = (cos(2x) + 1) / 2
->
sin^2 = (1 - cos(2x))/2

Answer 3

can you do the sin^2 part now ?

Answer 4

for the csc^2 you have to remember that the derivative of cot(x) is -csc^2
that should help you with the csc^2 part

Answer 5

ok, thanks... i will post what i think is right

Answer 6

(2x - 2sin2x) - (-cotx/4)

Answer 7

is that right?^

Answer 8

(2x - sin2x) - (-cotx/4)

Answer 9

problem with the 2 in front of the sin(2x)

Answer 10

where does the 4 go?... because it's 4(1-cos2x/2)... so i thought that would simplify to 2- 2cos2x

Answer 11

?

Answer 12

integral of (2- 2cos2x) = 2x - (2/2) * sin(2x) = 2x - sin(2x)