`int(cos 4x-1)/(cot x-tanx)dx` is equal to

Options

A. `k=-1//2`
B. `k=-1//8`
C. `k=-1//4`
D. None of these

Answer

B

Explanation

Let `l=int(cos 4x+1)/(cot x-tanx)dx=int(2 cos^(2)2x)/((cos^(2)x-sin^(2)x)/(sinx cosx))dx`
`=int(2cos^(2)2x)/(cos2x).sinx cosxdx`
`=int cos2x.sin 2x dx`
`=(1)/(2)int 2 cos 2x sin 2x dx`
`=(1)/(2)int sin 4x dx=(1)/(2).((-cos 4x))/(4)+C`
`rArr" "l=(-cos 4x)/(8)+C=k cos 4x+C" "therefore" "k=-(1)/(8)`