Physics, asked by JamesBond07, 1 year ago

Evaluate the integral, x = / ( 1 - sin wt) dt
,Where / = integration of
w = constant


MaheswariS: please write the question clearly

Answers

Answered by MaheswariS
55

Answer:

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

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}

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