# Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 28

$$\int \sec^2(12-25\theta) d\theta= -\frac{1}{25} \tan (12-25\theta)+c.$$

#### Work Step by Step

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

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