Re: Integral with sqrt(1 + t^2).

On Wed, 20 Jun 2007 18:11:42 -0700, mina_world@xxxxxxxxxxx wrote:

Hello sir~

int{0 to t} sqrt(1 + t^2) dt

= (1/2)[t.sqrt(1+t^2) + ln{t + sqrt(1+t^2)}].

--------------------------------------------------------
I want to show it.
so,
int{0 to t} sqrt(1 + t^2) dt = int{0 to t} (t)' . sqrt(1 + t^2) dt

= [t.sqrt(1+t^2)] - int{0 to t} t.{t / sqrt(1+t^2)} dt

I can't progress any more.
So, I need your advice.

Consider the substitution t = sinh(u)

You need to know some identities for the hyperbolic functions, and
also that

arcsinh(z) = ln[ z + sqrt(1 + z^2) ]

.

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