`int (cosx-cos2x)/(1-cosx) dx`

Options

A. ` 2 sin x + x + C`
B. ` 2 sin x - x + C`
C. ` cos x + x + C`
D. ` 2 cos x + x + C`

Answer

A

Explanation

Let `I = int(cosx-cos2x)/(1-cosx)dx = int(2sin'(3x)/(2).sin'(x)/(2))/(1-1+2sin^(2)'x/2)dx`
` = 2int(sin'(3x)/(2).sin'(x)/(2))/(2sin^(2)'(x)/(2))dx = int(sin'(3x)/(2))/(sin'x/2)dx`
`= int(3sin'x/2-4sin^(3)'x/2)/(sin'x/2), [:' sin3x = 3sinx - 4sinx^(3)x]`
`=3intdx-4intsin^(2)'x/2dx=3intdx-4int(1-cosx)/(2)dx`
`= 3dx - 2intdx + 2 intcos x dx`
`= intdx + 2 intcosx dx = x + 2 sinx + C = 2 sin x + x + C`