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    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.
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    (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
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    (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
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    (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
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    Yup get it!

    Thanks for your help
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    (Original post by mathsRus)
    Yup get it!

    Thanks for your help
    my pleasure
 
 
 
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