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

Question

UpStudy AI · Accepted Answer

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:**

$$
\int \sin (\omega t + \varphi) \, dt = -\frac{1}{\omega} \cos(\omega t + \varphi) + C
$$
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.

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

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Calculus Questions