The value of integral ∫π/43π/4 x/(1+sinx)dx

We have,I=∫π/43π/4 x/(1+sinx)dx
Multiply and divide LHS with (1-sinx), we get
I=∫π/43π/4 x(1-sinx)/(1-sin2x)dx
I=∫π/43π/4 x(1-sinx)/cos2xdx
I=∫π/43π/4 x(sec2x-secxtanx)dx
I=∫π/43π/4 xsec2xdx-∫π/43π/4 xsecxtanxdx
Using integration by parts
I=[(xtanx)-∫1.tanxdx)]π/43π/4+[(xsecx)-∫1.secxdx)]π/43π/4
I=[(xtanx)-(log|secx|)]π/43π/4+[(xsecx)+(log|secx+tanx|]π/43π/4
I=[(3π/4)tan(3π/4)-ln|sec(3π/4)|] - [( π/4)tan(π/4)-ln|sec(π/4)|] + [(3π/4)sec(3π/4)+ln|sec(3π/4)+tan(3π/4)|] - [(π/4)sec(π/4)+ln|sec(π/4)+tan(π/4)|]
I = π2(√2+1)