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Log (secx + tan x ) / cos x dx = log ( Secy + tan y ) / cos y dy

Log (secx + tan x ) / cos x dx = log ( Secy + tan y ) / cos y dy

Grade:12th pass

2 Answers

Arun
25763 Points
3 years ago
if this question is to be proved then,
we know if variable is changed then, no difference is generated. 
Example- integral of f(x) dx= integral of f(y) dy
hence your question is proved
Vikas TU
14149 Points
3 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)

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