## Calculus (3rd Edition)

$$\int \sec^2(12-25\theta) d\theta= -\frac{1}{25} \tan (12-25\theta)+c.$$
Recall that $(\tan x)'=\sec^2 x$. Let $u=12-25\theta$, then $du=-25d\theta$. $$\int \sec^2(12-25\theta) d\theta=-\frac{1}{25}\int \sec^2udu\\ =-\frac{1}{25} \tan u+c=-\frac{1}{25} \tan (12-25\theta)+c.$$