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