Sqare root integral

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August 22nd 2008, 08:41 PM
MatteNoob

Sqare root integral

Hello.
This isn't in my curriculum so I've never seen it done before. When my book meets an integral like this in the various examples it refers to the calculator.

I'd like to see how it is done, so please solve it step by step if you want to. :]

$\frac 12 \int_0^{10\pi}\left(\sqrt{t^2 + 1}\right)\rm{d}t$
August 22nd 2008, 09:19 PM
kalagota

you can do the general substitution rule.. $\int u \ dv = uv - \int v \ du$

let
$u = \sqrt{t^2+1} \Rightarrow du = \frac{t}{\sqrt{t^2+1}}$
$dv = dt \Rightarrow v=t$

so, $\int {\sqrt{t^2+1}} \ dt = t\sqrt{t^2+1} - \int \frac{t^2}{\sqrt{t^2+1}} \ dt$

if you set $t = \tan u \Rightarrow dt = \sec^2 u \ du$

the integral on the right side changes to

$\int \frac{\tan^2 u}{\sec u} \ du = \int \frac{\sin^2 u}{\cos} \ du$ ...

**************************************************

(another solution)
or you can substitute $\tan u = t$ from the start.. so, $\sec^2 u \ du = dt$

$\int {\sqrt{t^2+1}} \ dt = \int \sec^3 u \ du$

and you use $\int \sec^n x \ dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} \int\sec^{n-2} x \ dx$

noting that $\int \sec x \ dx = \ln |\sec x + \tan x| + C$
August 22nd 2008, 09:26 PM
MatteNoob

Cool!
I see you use trigonometric substitution in both solutions. I will try to find some resources on that topic.

Thanks a million for the solution (and your time), this seems to be a powerful way to solve many different integrals :)
August 22nd 2008, 10:06 PM
Marine

Hi there!

I will suggest another solution:

$\displaystyle{\int}\sqrt{t^2+1}dt$

$\cosh^2z-\sinh^2z=1$
$\cosh^2z=\sinh^2z+1$

let $t=\sinh z$
$dt=\cosh zdz$

$\displaystyle{\int}\sqrt{t^2+1}dt=\displaystyle{\i nt}\cosh^2zdz$

$\displaystyle{\int}\cosh^2zdz=\sinh z\cosh z-\displaystyle{\int}\sinh^2zdz=\sinh z\cosh z-\displaystyle{\int}(\cosh^2z-1)dz$

$\displaystyle{\int}\cosh^2zdz=\sinh z\cosh z-\displaystyle{\int}\cosh^2zdz+z$

$2\displaystyle{\int}\cosh^2zdz=\sinh z\cosh z+z$

$\displaystyle{\int}\cosh^2zdz=\frac{\sinh z\cosh z+z}{2}+C$

resubstitution: $t=\sinh z$ , $\cosh z=\sqrt{1+\sinh^2z}$ and $z={\rm}{arsinh}(t)$

or:

$\displaystyle{\int}\cosh^2zdz=\frac{\sinh z\cosh z+z}{2}+C=\frac{t\sqrt{1+t^2}+{\rm}{arsinh}(t)}{2} +C$

So, hyp substitutions are also helpful, to me even easier to use :)
August 22nd 2008, 11:36 PM
flyingsquirrel

Hello,

Quote:

Originally Posted by MatteNoob

I'd like to see how it is done, so please solve it step by step if you want to. :]

$\frac 12 \int_0^{10\pi}\left(\sqrt{t^2 + 1}\right)\rm{d}t$

Another approach, without substitution :
$\sqrt{t^2+1}=\sqrt{t^2+1}\cdot \frac{\sqrt{t^2+1}}{\sqrt{t^2+1}}=\frac{t^2+1}{\sq rt{t^2+1}}=\frac{t^2}{\sqrt{t^2+1}}+\frac{1}{\sqrt {t^2+1}}$

hence

$\begin{aligned}I=\frac 12 \int_0^{10\pi}\sqrt{t^2 + 1}\,\rm{d}t<br /> &=\frac 12 \int_0^{10\pi}\frac{t^2}{\sqrt{t^2 + 1}}\,\mathrm{d}t+\frac 12 \int_0^{10\pi}\frac{1}{\sqrt{t^2 + 1}}\,\mathrm{d}t\\<br /> &=\frac 12 \int_0^{10\pi}\frac{t^2}{\sqrt{t^2 + 1}}\,\mathrm{d}t+\frac{\mathrm{arcsinh} (10\pi)}{2}\end{aligned}$

Using integration by parts, ( $u=t$ and $v'=\frac{t}{\sqrt{t^2+1}}$ ) :

$\int \frac{t^2}{\sqrt{t^2 + 1}}\,\mathrm{d}t=t\sqrt{t^2+1}-\int \sqrt{t^2+1}\,\mathrm{d}t$

hence

$I=\frac{t\sqrt{t^2+1}}{2} \Big{|}_0^{10\pi} - \underbrace{\frac 12\int_0^{10\pi} \sqrt{t^2+1}\,\mathrm{d}t}_{I}+\frac{\mathrm{arcsi nh} (10\pi)}{2}$

$\boxed{I=\frac{10\pi \sqrt{100\pi^2+1}+\mathrm{arcsinh} (10\pi)}{4}}$