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24. \( \int e^{\cos (t)} \sin (t) d t \)

Ask by Erickson Boyd. in Spain
Feb 04,2025

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Answer

La integral \( \int e^{\cos(t)} \sin(t) \, dt \) se resuelve reemplazando \( \cos(t) \) con \( u \) y \( \sin(t) \, dt \) con \( -du \). Al integrar \( e^{u} \), obtenemos \( -e^{u} + C \). Sustituyendo de nuevo \( u \) por \( \cos(t) \), la solución es \( -e^{\cos(t)} + C \).

Solution

Para resolver la integral \[ \int e^{\cos(t)} \sin(t) \, dt, \] seguimos los siguientes pasos: 1. **Sustitución**: Sea \( u = \cos(t) \). Entonces, la derivada de \( u \) respecto a \( t \) es: \[ \frac{du}{dt} = -\sin(t) \quad \Rightarrow \quad du = -\sin(t) \, dt \] De aquí, podemos expresar \( \sin(t) \, dt \) en términos de \( du \): \[ \sin(t) \, dt = -du \] 2. **Reescribir la integral**: Sustituyendo en la integral original: \[ \int e^{\cos(t)} \sin(t) \, dt = \int e^{u} (-du) = -\int e^{u} \, du \] 3. **Integrar**: La integral de \( e^{u} \) respecto a \( u \) es \( e^{u} \). Por lo tanto: \[ -\int e^{u} \, du = -e^{u} + C \] donde \( C \) es la constante de integración. 4. **Deshacer la sustitución**: Reemplazamos \( u \) por \( \cos(t) \): \[ -e^{u} + C = -e^{\cos(t)} + C \] **Resultado final:** \[ \int e^{\cos(t)} \sin(t) \, dt = -e^{\cos(t)} + C \] donde \( C \) es la constante de integración.

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Mind Expander

To solve the integral \( \int e^{\cos (t)} \sin (t) d t \), we can use a substitution method. Let's set \( u = \cos(t) \), which means \( du = -\sin(t) dt \) or \( dt = -\frac{du}{\sin(t)} \). By changing the variable and manipulating the integral, we can express it in terms of \( u \): \[ \int e^{\cos(t)} \sin(t) dt = -\int e^u du = -e^u + C \] Now substituting back for \( u \): \[ -e^{\cos(t)} + C \] This gives us the final result of the integral: \[ \int e^{\cos(t)} \sin(t) dt = -e^{\cos(t)} + C \] where \( C \) is the constant of integration. Happy integrating!

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