## Calculus: Early Transcendentals 8th Edition

$$\frac{2}{\pi}$$
By substitution: Let $u = \frac{\pi}{2}t$. Hence, $du = \frac{\pi}{2}dt$. Thus, substituting $\frac{\pi}{2}t$ with $u$ and $dt$ with $\frac{2}{\pi}du$: $\int^1_0\cos({\frac{\pi}{2}t})dt$ $=\int^{\frac{\pi}{2}}_0\cos u (\frac{2}{\pi}du)$ $=\frac{2}{\pi}\int^{\frac{\pi}{2}}_0\cos u du$ $=\frac{2}{\pi}[\sin u]|^{\frac{\pi}{2}}_0$ $=\frac{2}{\pi}[\sin(\frac{\pi}{2})-\sin 0]$ $=\frac{2}{\pi}[1-0]$ $=\frac{2}{\pi}$