1. ## Integration of sqrt(1+t^2)dt

How can I integrate: $\int^1_{0}\sqrt{1+t^2}dt$?

The obvious choice seems like substitution, but I got a bit stuck doing it that way:

$u=1+t^2$
$\frac{du}{dt}=2t$
$du=2tdt$

If it were $du=2dt$, then I could just multiply the limits by 2, but since there is a $t$ in there too, I'm not sure what to do, since it leaves the expression still with the unknown, $t$, in there.

How can I do this?

Thanks

2. Originally Posted by Greengoblin
How can I integrate: $\int^1_{0}\sqrt{1+t^2}dt$?

The obvious choice seems like substitution, but I got a bit stuck doing it that way:

$u=1+t^2$
$\frac{du}{dt}=2t$
$du=2tdt$

If it were $du=2dt$, then I could just multiply the limits by 2, but since there is a $t$ in there too, I'm not sure what to do, since it leaves the expression still with the unknown, $t$, in there.

How can I do this?

Thanks
Thats exactly why its not a good idea to sub $u = 1+t^2$!

See the problem is the square root sign. We need to get rid of it... that is why choose a substitution that will get rid of it...

So sub $t = \tan u \Rightarrow dt = \sec^2 u \, du$,

$\int^1_{0}\sqrt{1+t^2}dt =\int^{\frac{\pi}4}_{0}\sec^3 u\, du$

Now try again

Note that you can also try the hyperbolic substitution $t = \sinh u$.

3. Hi
Originally Posted by Greengoblin
How can I integrate: $\int^1_{0}\sqrt{1+t^2}dt$?

The obvious choice seems like substitution, but I got a bit stuck doing it that way:

$u=1+t^2$
$\frac{du}{dt}=2t$
$du=2tdt$

If it were $du=2dt$, then I could just multiply the limits by 2, but since there is a $t$ in there too, I'm not sure what to do, since it leaves the expression still with the unknown, $t$, in there.
That's not really a problem : we also know that $u=1+t^2$ hence $t^2=u-1\implies t=\pm\sqrt{u-1}$. As $t\in[0,1]$, we can safely choose $t=+\sqrt{u-1}$. Substituting this in $\mathrm{d}u=2t\mathrm{d}t$, we get $\mathrm{d}u=2\sqrt{u-1}\mathrm{d}t$ that is to say $\mathrm{d}t=\frac{\mathrm{d}u}{2\sqrt{u-1}}$. (Note that the easiest way to find this is to write $t=\sqrt{u-1}$ and to differentiate with respect to $u$.)

Using this substitution, the original integral becomes $\int_1^2\frac{u}{2\sqrt{u-1}}\,\mathrm{d}u$... which is not easier to evaluate than the previous one.

4. Well if it is no easier, then I'll go for the first method mentioned.

Can you please explain where the $tanu$ came form?

5. Originally Posted by Greengoblin
Well if it is no easier, then I'll go for the first method mentioned.

Can you please explain where the $tanu$ came form?
As I mentioned previously, we have to get rid of the square root. That means we need to substitute such that 1+t^2 is some well known square function.

I remember the identities $1+\tan^2 u = \sec^2 u$ and $1+ \sinh^2 u = \cosh^2 u$. These identities suggest that I put $t = \tan u$ or $\sinh u$

6. Originally Posted by Greengoblin
How can I integrate: $\int^1_{0}\sqrt{1+t^2}dt$?

The obvious choice seems like substitution, but I got a bit stuck doing it that way:

$u=1+t^2$
$\frac{du}{dt}=2t$
$du=2tdt$

If it were $du=2dt$, then I could just multiply the limits by 2, but since there is a $t$ in there too, I'm not sure what to do, since it leaves the expression still with the unknown, $t$, in there.

How can I do this?

Thanks
Have you considered a hyperbolic substitution?

$t=\sinh(u)$

Then:

$\int^1_{0}\sqrt{1+t^2}~dt=\int_0^{\sinh^{-1}(1) }\cosh^2(u) ~du$

RonL

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# integral of sqrt(t^2 1-t)

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