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