Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 31

$$\int\frac{\sin(2t+1)}{\cos^2(2t+1)}dt=\frac{1}{2\cos(2t+1)}+C$$

$$A=\int\frac{\sin(2t+1)}{\cos^2(2t+1)}dt$$ We set $u=\cos(2t+1)$ Then $$du=(2t+1)'(-\sin(2t+1))dt=-2\sin(2t+1)dt$$ That means $$\sin(2t+1)dt=-\frac{1}{2}du$$ Therefore, $$A=-\frac{1}{2}\int\frac{1}{u^2}du=\frac{1}{2}\int-\frac{1}{u^2}du$$ $$A=\frac{1}{2}\times\frac{1}{u}+C=\frac{1}{2u}+C$$ $$A=\frac{1}{2\cos(2t+1)}+C$$