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