Very difficult integral to solve

[tonit]

Homework Statement

hey there guys, today I encountered a very difficult integral to solve, at least for me.
\int^\sqrt{3}_1 \frac{dx}{\sqrt{(1+x^2)^3}}

Homework Equations

The Attempt at a Solution

I've tried to substitute but didn't give results. Any suggestions please :)

 

[Dick]

What substitution did you try? I'd recommend a trig substitution, like x=tan(t).

 

[tonit]

Hmmm, alright I'm trying it now.

 

[tonit]

I'm now stuck at this point:

\int^\frac{\pi}{3}_\frac{\pi}{4} \frac{dt}{cos^3(t)}

 

[Dick]

I'm now stuck at this point:

\int^\frac{\pi}{3}_\frac{\pi}{4} \frac{dt}{cos^3(t)}

You'll need to show how you got there. That's not right and I can't guess what you did.

 

[tonit]

x = tan(t)
dx = d(tan[t]) = \frac{dt}{cos^2(t)}
x = 1 => tan(t) = 1 => t = \frac{\pi}{4}
x = \sqrt{3} => tan(t) = \sqrt{3} => t = \frac{\pi}{3}

so the integral becomes
\int^\frac{\pi}{3}_\frac{\pi}{4} \frac{dt}{cos^2(t)\sqrt{(1+tan^2(t))^3}}

oh and now I spot my mistake. It will become cos(t)dt, right?

 

[Dick]

x = tan(t)
dx = d(tan[t]) = \frac{dt}{cos^2(t)}
x = 1 => tan(t) = 1 => t = \frac{\pi}{4}
x = \sqrt{3} => tan(t) = \sqrt{3} => t = \frac{\pi}{3}

so the integral becomes
\int^\frac{\pi}{3}_\frac{\pi}{4} \frac{dt}{cos^2(t)\sqrt{(1+tan^2(t))^3}}

oh and now I spot my mistake. It will become cos(t)dt

Yes it will.

 

[tonit]

Ok, thanks for helping me :D