Question

Evaluate the integral, x= (1-sin omega t) dt with upper limit =π/omega and lower limit = 0​

Answer 1

Answer:

$x=\int\limits^{\frac{\pi}{\omega}}_0 (1-sin\omega{t}) \:dt=\frac{\pi-2}{\omega}$

Step-by-step explanation:

$x=\int\limits^{\frac{\pi}{\omega}}_0 (1-sin\omega{t}) \:dt$

$x=[t+\frac{cos\omega{t}}{\omega}]^{\frac{\pi}{\omega}}_0$

$x=[\frac{\pi}{\omega}+\frac{cos\omega(\frac{\pi}{\omega})}{\omega}]-[0+\frac{cos\omega{0}}{\omega}]$

$x=[\frac{\pi}{\omega}+\frac{cos\pi}{\omega}]-[0+\frac{cos\:0}{\omega}]$

$x=[\frac{\pi}{\omega}+\frac{(-1)}{\omega}]-[0+\frac{1}{\omega}]$

$x=\frac{\pi}{\omega}-\frac{1}{\omega}-\frac{1}{\omega}$

$x=\frac{\pi}{\omega}-\frac{2}{\omega}$

$x=\frac{\pi-2}{\omega}$

Answer 2

hope this helps you!!!!!!!!!!