The correct option is B −xcotx2+C
∫x−sinx1−cosxdx=∫x1−cosxdx−∫sinx1−cosxdx=12∫x cosec2(x2)dx−∫2sin(x2)cos(x2)2sin2(x2)dx⎡⎢
⎢
⎢
⎢⎣∵1−cosx=2sin2(x2)sinx=2sin(x2)⋅cos(x2)⎤⎥
⎥
⎥
⎥⎦=∫[x2⋅ cosec2(x2)+(−cot(x2))]dx
This is of the form ∫[f(x)+xf′(x)]dx
where
f(x)=−cot(x2);f′(x)=12 cosec2(x2)
We know that,
∫[f(x)+xf′(x)]dx=xf(x)+C
Thus,
∫x−sinx1−cosxdx=−xcot(x2)+C