Correct Answer - B
`I=int(cos^(3)x)/(sin^(2)x+sinx)dx`
`=int(cosx.(1-sin^(2)x))/(sinx(1+sinx))dx`
Put `sinx=t,` then `cos xdx =dt`
`rArr" "I=int((1-t)(1+t)dt)/(t(1+t))`
`=int((1-t)dt)/(t)`
`=int((1)/(t)-1)dt`
`=log_(e)|t|-t+C`
`=log_(e)|sinx|-sinx+C`