# Very difficult integral to solve

## Homework Statement

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

## The Attempt at a Solution

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

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

Hmmm, alright I'm trying it now.

I'm now stuck at this point:

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

Dick
Homework Helper
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.

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
Homework Helper
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.

Ok, thanks for helping me :D