Integral of $\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$ I am trying to find $$\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$$
$t = \sec \theta$ $dt = \sec \theta \tan\theta $
$$\int_\sqrt{2}^2 \frac{dt}{\sec ^2 \theta \sqrt{\sec^2 \theta-1}}$$
$$\int_\sqrt{2}^2 \frac{dt}{\sec ^2 \theta \tan^2 \theta}$$
$$\int_\sqrt{2}^2 \frac{\sec \theta \tan\theta}{\sec ^2 \theta \tan^2 \theta}$$
$$\int_\sqrt{2}^2 \frac{1}{\sec \theta}$$
$$\int_\sqrt{2}^2 \cos \theta$$
$$\sin \theta$$
Then I need to make it in terms of t.
$t = \sec \theta$
So I just use the arcsec which is
$\theta =\operatorname{arcsec} t$
$$\sin (\operatorname{arcsec} t)$$
This is wrong but I am not sure why.

$$\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$$
Substitute :
$t = \sec \theta$
$dt = \sec \theta \tan\theta d\theta$
The limits will also change accordingly
When $t=2$ , $\theta = \ arc sec(2) = \frac{\pi}{3}$
When $t=\sqrt2$ , $\theta = \ arc sec(\sqrt2) = \frac{\pi}{4}$
$$=\int_\frac{\pi}{4}^\frac{\pi}{3} \frac{\sec \theta \tan\theta d\theta}{\sec ^2 \theta \tan \theta}$$
$$=\int_\frac{\pi}{4}^\frac{\pi}{3} \cos \theta d \theta$$
$$=\sin \frac{\pi}{3} - \sin \frac{\pi}{4}$$
$$=\frac{\sqrt3 - \sqrt2}{2}$$

I think the answer you got is also correct
$$\sin (\operatorname{arcsec} (t))$$
by applying simple trigonometric rules
$$=\frac{\sqrt{t^2-1}}{t}$$
and then applying the limits we get the same answer
$$=\frac{\sqrt3 - \sqrt2}{2}$$ 
