Integration problem: int [x=0,1] [sqrt(1+x^2)] dx

[kysmalte]

I need to calculate the integral from 0 to 1 of sqrt(1+x^2). I know how to do definite integrals but I cannot resolve the square root part.

Any help is greatly appreciated, thanks!

 

[morson]

Try making the substitution x = Tan(A). You should get a trig function, which you could probably integrate by parts.

 

[kysmalte]

When I do that I end up with sqrt(1+tan^2(a)) which becomes sqrt(sec^2(a)) which reduces to sec(a). x=tan(a), dx=sec^2(a) da. The entire integral becomes the integral of sec^3(a) da. I tried to break this up into sec(a)sec^2(a) and run integration by parts but I couldn't arrive at an answer. Any suggestions?

 

[morson]

It's a matter of breaking the integral down (which may mean using integration by parts
more than once). I get:

\(\displaystyle \L\\\int{sec^3(a)}da \:=\:\frac{1}{2}\\)seca*tana + \(\displaystyle \frac{1}{2}\\)loge(seca + tana) + \(\displaystyle C\)

But since this is a definite integral, the C is no longer needed. Don't forget to update the limits.

 

[Count Iblis]

I need to calculate the integral from 0 to 1 of sqrt(1+x^2). I know how to do definite integrals but I cannot resolve the square root part.

Any help is greatly appreciated, thanks!

You can also substitute x = sinh(t). Then sqrt(1+x^2) = cosh(t) and
dx = cosh(t) dt, so you have to integrate cosh^2(t), which is trivial, e.g. you can just expand out the square of exp(t) + exp(-t) .

 

[galactus]

You can try to derive the general case and then use n=3.

\(\displaystyle \L\\\int{sec^{n}(x)dx=\frac{1}{n-1}sec^{n-2}(x)tan(x)+\frac{n-2}{n-1}\int{sec^{n-2}(x)}dx\)

Write $\displaystyle sec^{n}(x)=sec^{n-2}(x)sec^{2}(x)$, intgerate by parts with

$\displaystyle u=sec^{n-2}(x)$ and $\displaystyle dv=sec^{2}(x)dx$, and use the trig formula $\displaystyle tan^{2}(x)+1=sec^{2}(x)$

To the derive the case when n=3, as you have, let $\displaystyle u=sec(x), \;\ dv=sec^{2}(x)dx, \;\ du=sec(x)tan(x)dx, \;\ v=tan(x)$