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(1) \( \int \sin (\omega t+\varphi) \mathrm{d} t(\omega, \varphi \quad \) are constants \( ) \)

Ask by Love Cervantes. in China
Jan 23,2025

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Answer

The integral of \( \sin (\omega t + \varphi) \) with respect to \( t \) is \( -\frac{1}{\omega} \cos(\omega t + \varphi) + C \), where \( C \) is the constant of integration.

Solution

To evaluate the integral: \[ \int \sin (\omega t + \varphi) \, dt \] where \( \omega \) and \( \varphi \) are constants, you can use a basic integration rule for sine functions. The integral of \( \sin(k t + c) \) with respect to \( t \) is: \[ -\frac{1}{k} \cos(k t + c) + C \] Applying this to your specific integral: \[ \int \sin (\omega t + \varphi) \, dt = -\frac{1}{\omega} \cos(\omega t + \varphi) + C \] where \( C \) is the constant of integration. **Final Answer:** \[ \int \sin (\omega t + \varphi) \, dt = -\frac{1}{\omega} \cos(\omega t + \varphi) + C \]

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To integrate \( \sin (\omega t + \varphi) \), we can use a simple substitution method. The integral becomes: \[ \int \sin(\omega t + \varphi) \, \mathrm{d}t = -\frac{1}{\omega} \cos(\omega t + \varphi) + C \] where \( C \) is the constant of integration. This means you'll get a negative cosine function scaled by the constant \( \frac{1}{\omega} \). This integration is widely used in physics and engineering, especially in oscillatory motion problems like analyzing waves, pendulums, and other periodic phenomena. Understanding this integration can help in modeling real-world systems, such as electrical circuits and mechanical vibrations, making it a valuable tool for students and professionals alike!

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