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Encontrar la integral \( \int \sin ^{5} 2 x \cos 2 x d x \)

Ask by Gough Peterson. in Mexico
Mar 13,2025

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Answer

La integral \( \int \sin^5(2x) \cos(2x) \, dx \) se resuelve como \( \frac{\sin^6(2x)}{12} + C \), donde \( C \) es una constante.

Solution

Consideremos la integral \[ \int \sin^5(2x) \cos(2x) \, dx. \] **Paso 1.** Realizamos el cambio de variable: \[ u = \sin(2x). \] Calculamos la derivada de \( u \): \[ \frac{du}{dx} = 2 \cos(2x) \quad \Longrightarrow \quad du = 2 \cos(2x) \, dx. \] De aquí se deduce que: \[ \cos(2x) \, dx = \frac{du}{2}. \] **Paso 2.** Sustituimos en la integral: \[ \int \sin^5(2x) \cos(2x) \, dx = \int u^5 \frac{du}{2} = \frac{1}{2} \int u^5 \, du. \] **Paso 3.** Integramos con respecto a \( u \): \[ \int u^5 \, du = \frac{u^6}{6}. \] Entonces, la integral se vuelve: \[ \frac{1}{2} \cdot \frac{u^6}{6} = \frac{u^6}{12}. \] **Paso 4.** Regresamos a la variable original usando \( u = \sin(2x) \): \[ \frac{u^6}{12} = \frac{\sin^6 (2x)}{12}. \] Por lo tanto, la solución de la integral es: \[ \int \sin^5(2x) \cos(2x) \, dx = \frac{\sin^6(2x)}{12} + C, \] donde \( C \) es la constante de integración.

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Para resolver la integral \( \int \sin^{5}(2x) \cos(2x) \, dx \), podemos usar un cambio de variable. Observamos que la derivada de \(\sin(2x)\) es \(2\cos(2x)\), por lo que podemos hacer el reemplazo: Sea \( u = \sin(2x) \). Entonces, \( du = 2\cos(2x) \, dx \) o \( \frac{1}{2} du = \cos(2x) \, dx \). Reescribimos la integral como: \[ \int \sin^{5}(2x) \cos(2x) \, dx = \int u^{5} \frac{1}{2} \, du = \frac{1}{2} \int u^{5} \, du. \] Ahora resolvemos la integral: \[ \frac{1}{2} \left( \frac{u^{6}}{6} \right) + C = \frac{1}{12} u^{6} + C. \] Sustituyendo \( u = \sin(2x) \) de nuevo, obtenemos: \[ \frac{1}{12} \sin^{6}(2x) + C. \] Por lo tanto, la integral \(\int \sin^{5}(2x) \cos(2x) \, dx\) es: \[ \frac{1}{12} \sin^{6}(2x) + C. \]

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