How integrate

∫ sqrt (1+x²)

On Monday, 25 April 2022 at 09:01:34 UTC-3, Richard Hachel wrote:
> How integrate
>
> ∫ sqrt (1+x²)

Use the trigonometric substitution x = tan t, dx = sec^2 t, t = arctan x (with t in (-pi/2, +pi/2) )

That gives ∫ sqrt (1+x²) dx = ∫ sec^3 t dt, which you can integrate by parts. It's not pretty, but it leads to a closed-form solution.

Richard Hachel a écrit :
> How integrate
>
> ∫ sqrt (1+x²)

In[1]:= Integrate[Sqrt[1 + x^2], x]
Out[1]= 1/2 (x Sqrt[1 + x^2] + ArcSinh[x])

Le 25/04/2022 à 14:47, Gus Gassmann a écrit :
> On Monday, 25 April 2022 at 09:01:34 UTC-3, Richard Hachel wrote:
>> How integrate
>>
>> ∫ sqrt (1+x²)
>
> Use the trigonometric substitution x = tan t, dx = sec^2 t, t = arctan x (with t
> in (-pi/2, +pi/2) )
>
> That gives ∫ sqrt (1+x²) dx = ∫ sec^3 t dt, which you can integrate by
> parts. It's not pretty, but it leads to a closed-form solution.

Merci beaucoup.

R.H.

Le 25/04/2022 à 14:47, Gus Gassmann a écrit :
> On Monday, 25 April 2022 at 09:01:34 UTC-3, Richard Hachel wrote:
>> How integrate
>>
>> ∫ sqrt (1+x²)
>
> Use the trigonometric substitution x = tan t, dx = sec^2 t, t = arctan x (with t
> in (-pi/2, +pi/2) )
>
> That gives ∫ sqrt (1+x²) dx = ∫ sec^3 t dt, which you can integrate by
> parts. It's not pretty, but it leads to a closed-form solution.

x= 0 to 5.0245
∫ sqrt (1+x²)

? ? ?
R.H.

Richard Hachel wrote:

>> On Monday, 25 April 2022 at 09:01:34 UTC-3, Richard Hachel wrote:
>>> How integrate ∫ sqrt (1+x²)
>>
>> Use the trigonometric substitution x = tan t, dx = sec^2 t, t = arctan
>> x (with t in (-pi/2, +pi/2) )
>>
>> That gives ∫ sqrt (1+x²) dx = ∫ sec^3 t dt, which you can integrate by
>> parts. It's not pretty, but it leads to a closed-form solution.
>
> x= 0 to 5.0245
>
> ∫ sqrt (1+x²)
> ? ? ?

========================
Integration indefinite
========================
________
╱ 2
x⋅╲╱ x + 1 asinh(x)
───────────── + ────────
2 2
========================
Integration no sym 0.0 to 5.0245
========================
n = 14.028988247899438e+000

Le 25/04/2022 à 22:43, Dean Totolos a écrit :
> Richard Hachel wrote:
>
>>> On Monday, 25 April 2022 at 09:01:34 UTC-3, Richard Hachel wrote:
>>>> How integrate ∫ sqrt (1+x²)
>>>
>>> Use the trigonometric substitution x = tan t, dx = sec^2 t, t = arctan
>>> x (with t in (-pi/2, +pi/2) )
>>>
>>> That gives ∫ sqrt (1+x²) dx = ∫ sec^3 t dt, which you can integrate by
>>> parts. It's not pretty, but it leads to a closed-form solution.
>>
>> x= 0 to 5.0245
>>
>> ∫ sqrt (1+x²)
>> ? ? ?
>
> ========================
> Integration indefinite
> ========================
> ________
> ╱ 2
> x⋅╲╱ x + 1 asinh(x)
> ───────────── + ────────
> 2 2
> ========================
> Integration no sym 0.0 to 5.0245
> ========================
> n = 14.028988247899438e+000

Merci beaucoup.

R.H.

Richard Hachel wrote:

>> ======================== Integration indefinite
>> ========================
>> ________
>> ╱ 2
>> x⋅╲╱ x + 1 asinh(x) ───────────── + ────────
>> 2 2
>> ======================== Integration no sym 0.0 to 5.0245
>> ========================
>> n = 14.028988247899438e+000
>
> Merci beaucoup.

no problema, my friend. Otherwise the "ukrainian" nazis are throwing mines
in Black Sea, so it can't be the russians. Hence if something hit that
ship from outside, was a mine they couldn't see, not a "ukrainian missile"
as the nazi western mass_media says. What a wankers, the western
mass_media.

Le 25/04/2022 à 23:06, Dean Totolos a écrit :
> no problema, my friend. Otherwise the "ukrainian" nazis are throwing mines
> in Black Sea, so it can't be the russians. Hence if something hit that
> ship from outside, was a mine they couldn't see, not a "ukrainian missile"
> as the nazi western mass_media says. What a wankers, the western
> mass_media.

Westerners do not want to find a peaceful settlement, of course.

They want a world war.

I will never believe that we have been pissing everyone off for three
years for a little flu.

All these bites, these masks, these restrictions, these precautions,
etc... It makes me think of the preparations for a bacteriological world
war.

R.H.

Le 25/04/2022 à 14:01, Richard Hachel a écrit :
> How integrate
>
> ∫ sqrt (1+x²)

And now?

∫ sqrt (1+ax²)

R.H.

Richard Hachel wrote:

> Le 25/04/2022 à 14:01, Richard Hachel a écrit :
>> How integrate ∫ sqrt (1+x²)
>
> And now? ∫ sqrt (1+ax²)

__________
╱ 2
x⋅╲╱ a⋅x + 1 asinh(√a⋅x)
─────────────── + ───────────
2 2⋅√a

Le 26/04/2022 à 18:00, Abe Banos a écrit :
> Richard Hachel wrote:
>
>> Le 25/04/2022 à 14:01, Richard Hachel a écrit :
>>> How integrate ∫ sqrt (1+x²)
>>
>> And now? ∫ sqrt (1+ax²)
>
> __________
> ╱ 2
> x⋅╲╱ a⋅x + 1 asinh(√a⋅x)
> ─────────────── +
> ───────────
> 2 2⋅√a

Merci.

R.H.