Vikas TU
Last Activity: 7 Years ago
Dear Student,
It can be written as:
log(secx + tanx)dx/cosx = log(secy + tany)dy/cosy
=> secxlog(secx + tanx)dx = secylog(secy + tany)dy
let u = secx. du = log(secx + tanx)dx. Similarly v = = secy. dv = log(secy + tany)dy.
=> udu = vdv.
Integrating both sides, we get u2/2 = v2/2 + c/2.
i.e. sec2x = sec2y + c.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)