You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Integration problem: int [x=0,1] [sqrt(1+x^2)] dx
- Thread starter kysmalte
- Start date
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?
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.
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
Junior Member
- Joined
- Feb 13, 2007
- Messages
- 108
kysmalte said: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) .
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 \(\displaystyle u=sec(x), \;\ dv=sec^{2}(x)dx, \;\ du=sec(x)tan(x)dx, \;\ v=tan(x)\)
\(\displaystyle \L\\\int{sec^{n}(x)dx=\frac{1}{n-1}sec^{n-2}(x)tan(x)+\frac{n-2}{n-1}\int{sec^{n-2}(x)}dx\)
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 \(\displaystyle u=sec(x), \;\ dv=sec^{2}(x)dx, \;\ du=sec(x)tan(x)dx, \;\ v=tan(x)\)