If `I= int (sin 2x)/((3+4cosx)^(3))dx,` then `I` equals

Options

A. `(3cos x+8)/((3+4cosx)^(2))+C`
B. `(3+8cos x)/(16(3+4cosx)^(2))+C`
C. `(3+cos x)/((3+4cosx)^(2))+C`
D. `(3-8cos x)/(16(3+4cosx)^(2))+C`

Answer

B

Explanation

`I= int (sin 2x)/((3+4cosx)^(3))dx`
Put `3+4cosx=t," so that " -4sinx dx=dt.` Then
`I=(-1)/(8)int((t-3))/(t^(3))dt=(1)/(8)((1)/(t)-(3)/(2)(1)/(t^(2)))+C`
`=(2t-3)/(16t^(2))=(8cosx+3)/(16(3+4cosx)^(2))+C`