I have to integrate int_0^1 sqrt(t^2-1)dt

[Jonny_trigonometry]

I was wondering how to solve this integral:

[tex]\int_{0}^{1}\sqrt{t^2-1}\,dt[/tex]

When I punch it into mathematica, it gives:

[tex] 1/2 t\sqrt{-1+t^2}-1/2\log{(t+\sqrt{-1+t^2})} [/tex]

I was wondering what steps are done to get this result

I suppose I forgot to enter it in as a definate integral, but still...

 

[benorin]

I was wondering how to solve this integral:

[tex]\int_{0}^{1}\sqrt{t^2-1}\,dt[/tex]

When I punch it into mathematica, it gives:

[tex] 1/2 t\sqrt{-1+t^2}-1/2\log{(t+\sqrt{-1+t^2})} [/tex]

I was wondering what steps are done to get this result

Note that I used maple to find this substitution.

Start with [tex]\int\sqrt{t^2-1}\,dt[/tex] and apply the substitution [tex]t=\sec(\theta)\Rightarrow dt=\tan(\theta)\sec(\theta)d\theta[/tex] which gives [tex]\sqrt{t^2-1}=\tan(\theta)[/tex] so that the integral becomes [tex]\int\tan^{2}(\theta)\sec(\theta)d\theta[/tex], you should be able to work it from there.

 

[VietDao29]

Or you can try another way:
Let:
[tex]\sqrt{t ^ 2 - 1} = x - t[/tex]. Differentiate both sides gives:
[tex]\frac{t dt}{\sqrt{t ^ 2 - 1}} = dx - dt[/tex]
[tex]\Leftrightarrow \left( \frac{t}{\sqrt{t ^ 2 - 1}} + 1 \right) dt = dx[/tex]
[tex]\Leftrightarrow \left( \frac{t + \sqrt{t ^ 2 - 1}}{\sqrt{t ^ 2 - 1}} \right) dt = dx[/tex]
But you should note that:
[tex]\sqrt{t ^ 2 - 1} + t = x[/tex]
That gives:
[tex]\Leftrightarrow \left( \frac{x}{\sqrt{t ^ 2 - 1}} \right) dt = dx[/tex]
[tex]\Leftrightarrow \frac{dt}{\sqrt{t ^ 2 - 1}} = \frac{dx}{x}[/tex]
Integrate both sides gives:
[tex]\int \frac{dt}{\sqrt{t ^ 2 - 1}} = \int \frac{dx}{x} = \ln x = \ln (\sqrt{t ^ 2 - 1} + t)[/tex].
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In general, you can show that:
[tex]\int \frac{dt}{\sqrt{t ^ 2 + \alpha}} = \ln (\sqrt{t ^ 2 + \alpha} + t)[/tex].
-----------------
You can try to integrate [tex]\int \sqrt{t ^ 2 - 1} dt[/tex] by parts, then you may need to use [tex]\int \frac{dt}{\sqrt{t ^ 2 + \alpha}} = \ln (\sqrt{t ^ 2 + \alpha} + t)[/tex].
Viet Dao,

 

[arildno]

The simplest is possibly to use the substitution [itex]t=Cosh(u)[/itex]
However, your limits can't be right, unless you're dealing with a complex integral, rather than a real one.

 

[Jonny_trigonometry]

hmm... all these help a lot, thanks people!