Hyperbolic Calculus

Differentiate arccosh(secx)

and hence show (using ln form of a suitable inverse hyperbolic function) that for 0<=x<pi/2 we have

integral (secx)dx = ln(secx + tanx)

For the differentiation part I got secx as my answer which I do not know whether it is correct, but its the next part I have no idea.

Reply 1

TeeEm

19

Original post by mathsRus

Differentiate arccosh(secx)

and hence show (using ln form of a suitable inverse hyperbolic function) that for 0<=x<pi/2 we have

integral (secx)dx = ln(secx + tanx)

For the differentiation part I got secx as my answer which I do not know whether it is correct, but its the next part I have no idea.

part (a) is correct
d/dx[arcosh(secx)] = secx

integrate both sides w.r.t x gives

arcosh(secx)] = INT(secx) dx

do you know what the arcosh function is in logarithmic form?

apply to the left and you have the answer

Reply 2

mathsRus

OP

8

Original post by TeeEm

part (a) is correct
d/dx[arcosh(secx)] = secx

integrate both sides w.r.t x gives

arcosh(secx)] = INT(secx) dx

do you know what the arcosh function is in logarithmic form?

apply to the left and you have the answer

I found out arccosh in ln form: ln(x+sqrt(x^2 -1))

Why do I integrate both sides?

In other words, is the question trying to say that prove arccosh(secx) is the same as integral of secx?

I have some idea but really confused

Reply 3

TeeEm

19

Original post by mathsRus

I found out arccosh in ln form: ln(x+sqrt(x^2 -1))

Why do I integrate both sides?

In other words, is the question trying to say that prove arccosh(secx) is the same as integral of secx?

I have some idea but really confused

you integrate both sides because you try to prove if I understood well that

integral (secx)dx = ln(secx + tanx)

d/dx[arcosh(secx)] = secx

integrate w.r.t x

arcosh(secx) = integral (secx)dx

ln(secx+...) = integral (secx)dx

and the result trivially follows