The value of intlimits_0^{frac{pi }{{2omega } | vedclass

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(B) To evaluate the integral $I = \int\limits_0^{\frac{\pi }{{2\omega }}} {5\,\sin \omega t} \,dt$,we proceed as follows:First,pull the constant $5$ o

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$\frac{5}{\omega }$

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(B) To evaluate the integral $I = \int\limits_0^{\frac{\pi }{{2\omega }}} {5\,\sin \omega t} \,dt$,we proceed as follows:First,pull the constant $5$ out of the integral:$I = 5 \int\limits_0^{\frac{\pi }{{2\omega }}} \sin \omega t \,dt$Recall that the integral of $\sin \omega t$ with respect to $t$ is $-\frac{\cos \omega t}{\omega}$:$I = 5 \left[ -\frac{\cos \omega t}{\omega} \right]_0^{\frac{\pi }{{2\omega }}}$$I = -\frac{5}{\omega} \left[ \cos \omega t \right]_0^{\frac{\pi }{{2\omega }}}$Now,substitute the limits:$I = -\frac{5}{\omega} \left[ \cos \left( \omega \cdot \frac{\pi}{2\omega} \right) - \cos(0) \right]$$I = -\frac{5}{\omega} \left[ \cos \left( \frac{\pi}{2} \right) - \cos(0) \right]$Since $\cos \left( \frac{\pi}{2} \right) = 0$ and $\cos(0) = 1$:$I = -\frac{5}{\omega} [0 - 1]$$I = -\frac{5}{\omega} [-1]$$I = \frac{5}{\omega}$

The value of $\int\limits_0^{\frac{\pi }{{2\omega }}} {5\,\sin \omega t} \,dt$ is

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