Integral of $\int_\sqrt{2}^2 \frac{dt}{t^2 \sqrt{t^2-1}}$

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 $\displaystyle \int_{\pi/4}^{\pi/6} \frac{\sec\theta \tan\theta}{\sec^{2}\theta \cdot \tan\theta} \ d\theta = \int_{\pi/4}^{\pi/6} \cos\theta \ d\theta$.