## Calculus: Early Transcendentals 8th Edition

$$\int\cos(1+5t)dt=\frac{\sin(1+5t)}{5}+C$$
$$A=\int\cos(1+5t)dt$$ Let $u=1+5t$ Then we have $du=5dt$. That means $dt=\frac{1}{5}du$ Substitute into $A$: $$A=\frac{1}{5}\int\cos udu$$ $$A=\frac{1}{5}\sin u+C$$ $$A=\frac{\sin(1+5t)}{5}+C$$