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

• Jul 10th 2008, 11:58 PM
Greengoblin
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
• Jul 11th 2008, 12:05 AM
Isomorphism
Quote:

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 $t = \sinh u$.
• Jul 11th 2008, 12:16 AM
flyingsquirrel
Hi
Quote:

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.
• Jul 11th 2008, 12:22 AM
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?
• Jul 11th 2008, 12:53 AM
Isomorphism
Quote:

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 $\sinh u$
• Jul 11th 2008, 02:36 AM
CaptainBlack
Quote:

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