The value of ∫1+sinx1−sinxdx is
2tanx2+π4+C
2tanx2+π4+x+C
2tanx2+π4−x+C
2tan2x2+π4−x+C
∫1+sinx1−sinxdx=∫(1+sinx)21−sin2xdx=∫(secx+tanx)2dx=∫sec2x+∫sec2x−1dx+2∫secxtanxdx=2tanx−x+2secx+C=21+sinxcosx−x+C=21−cosx+π2sinx+π2−x+C=22sin2x2+π42sinx2+π4cosx2+π4−x+C=2tanx2+π4−x+C