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.
 A: These are the mistakes you have made:


*

*When you substitute $t = \sec\theta$ the limits will change accordingly.

*And $\sqrt{\sec^{2}\theta -1} \neq \tan^{2}\theta$ , its $\tan\theta$.

*When $t= \sec\theta$, $\theta$ will change from $\frac{\pi}{4}$ to $\frac{\pi}{3}$.

*When you make the change there will be a $d\theta$ term.

*Your integral will look like $\displaystyle \int_{\pi/4}^{\pi/6} \frac{\sec\theta \tan\theta}{\sec^{2}\theta \cdot \tan\theta} \ d\theta = \int_{\pi/4}^{\pi/6} \cos\theta \ d\theta$.
A: $$\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$$
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
$$=\frac{\sqrt3 - \sqrt2}{2}$$ 
A: To compute $$\int \frac{dt}{t^2 \sqrt{t^2-1}}$$ you may note that  $$\cosh^2(x)-\sinh^2(x)=1,$$ or $$\cosh^2(x)-1=\sinh^2(x),$$ so putting $t=\cosh(x),$ then $dt=\sinh(x)dx$
$$\int \frac{dt}{t^2 \sqrt{t^2-1}}=\int \frac{\sinh(x)dx}{\cosh^2(x)\sqrt{\cosh^2(x)-1}}$$ 
$$=\int \frac{\sinh(x)dx}{\cosh^2(x)\sqrt{\sinh^2(x)}}=\int\frac{dx}{\cosh^2(x)}$$
$$=\int{\text{sech}^2(x)dx}=\tanh(x)+C$$
and I leave the change of the bounds $t=\sqrt{2}$ to $t=2$ to you.
