## Calculus (3rd Edition)

$$\tan1.$$
Let $u=\cos \theta$, then $du=-\sin\theta d\theta$ and $u:1\to 0$. Now we have $$\int_{0}^{\pi / 2} \sec ^{2}(\cos \theta) \sin \theta d \theta=-\int_1^0\sec^2u du\\ =\int_0^1\sec^2u du=\tan u|_0^1=\tan1.$$