Đáp án: `bbB`
`I=int_0^(π/3) (1 + x sinx)/(cos^2x)dx`
`= int_0^(π/3) (1+xsinx) sec^2x dx`
`= int_0^(π/3) (sec^2x + x tanx secx) dx`
`= int_0^(π/3) sec^2x dx+int_0^(π/3) x tanx secx dx`
Đặt `{(u=x),(dv=tanxsecxdx):}=>{(du=dx),(v=secx):}`
Khi đó, `I=x secx |_0^(π/3) - int_0^(π/3) secx dx + int_0^(π/3) sec^2x dx`
`= (2 π)/3 -ln(tanx + secx) |_0^(π/3) + tanx|_0^(π/3)`
`=sqrt(3) + (2 π)/3 - ln(2 + sqrt(3))`
`=sqrt(3) + (2 π)/3 + ln(2 - sqrt(3))`
