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

[Jonny_trigonometry]

I was wondering how to solve this integral:

$$\int_{0}^{1}\sqrt{t^2-1}\,dt$$

When I punch it into mathematica, it gives:

$$1/2 t\sqrt{-1+t^2}-1/2\log{(t+\sqrt{-1+t^2})}$$

I was wondering what steps are done to get this result

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

 

[benorin]

I was wondering how to solve this integral:

$$\int_{0}^{1}\sqrt{t^2-1}\,dt$$

When I punch it into mathematica, it gives:

$$1/2 t\sqrt{-1+t^2}-1/2\log{(t+\sqrt{-1+t^2})}$$

I was wondering what steps are done to get this result

Note that I used maple to find this substitution.

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

 

[VietDao29]

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

 

[arildno]

The simplest is possibly to use the substitution $t=Cosh(u)$
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!