Question

очень срочно!!!
найдите интеграл используя метод интегрирования по частям ∫(5x+12)sinx/3dx​

Answer 1

Ответ:

Интеграл:

$\boldsymbol{\boxed{\int {(5x + 12) \sin \bigg( \frac{x}{3} \bigg) } \, dx = 45\sin\bigg(\dfrac{x}{3} \bigg)-15x\cos\bigg(\dfrac{x}{3} \bigg) -36\cos\bigg(\dfrac{x}{3} \bigg)+ C}}$

Примечание:

По таблице интегралов:

$\boxed{\int {\sin x} \, dx = -\cos x + C }$

$\boxed{\int {\cos x} \, dx = \sin x + C }$

Интегрирование по частям:

$\boxed{ \int {u} \, dv = uv - \int {v} \, du }$

По свойствам интегралов:

$\boxed{ \displaystyle \int \sum\limits_{i=1}^n {C_{i}f_{i}(x)} \, dx = \sum\limits_{i=1}^nC_{i} \int {f_{i}(x)} \, dx}$

Пошаговое объяснение:

$\displaystyle \int {(5x + 12) \sin \bigg( \frac{x}{3} \bigg) } \, dx = \int {\Bigg(5x\sin \bigg( \frac{x}{3} \bigg) + 12 \sin \bigg( \frac{x}{3} \bigg) \Bigg)} \, dx =$

$\displaystyle =\int {5x\sin \bigg( \frac{x}{3} \bigg) } \, dx + \int { 12 \sin \bigg( \frac{x}{3} \bigg) } \, dx =$

а)

$\displaystyle \int {5x\sin \bigg( \frac{x}{3} \bigg) } \, dx =$

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$u = 5x \Longrightarrow du = (5x)' \ dx = 5 \ dx$

$\displaystyle dv = \sin \bigg(\dfrac{x}{3} \bigg) \ dx \Longrightarrow v = \int{\sin \bigg(\dfrac{x}{3} \bigg)} \, dx = 3\int{\sin \bigg(\dfrac{x}{3} \bigg)} \, d \bigg(\dfrac{x}{3} \bigg) = -3\cos\bigg(\dfrac{x}{3} \bigg)$

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$\displaystyle = -15x\cos\bigg(\dfrac{x}{3} \bigg) - \int { -15\cos\bigg(\dfrac{x}{3} \bigg)} \, dx = 15\int {\cos\bigg(\dfrac{x}{3} \bigg)} \, dx -15x\cos\bigg(\dfrac{x}{3} \bigg)=$

$\displaystyle =(15 \cdot 3) \int {\cos\bigg(\dfrac{x}{3} \bigg)} \, d\bigg(\dfrac{x}{3} \bigg)-15x\cos\bigg(\dfrac{x}{3} \bigg)= 45\sin\bigg(\dfrac{x}{3} \bigg)-15x\cos\bigg(\dfrac{x}{3} \bigg) + C_{1}$

б)

$\displaystyle \int { 12 \sin \bigg( \frac{x}{3} \bigg) } \, dx = (12 \cdot 3)\int{\sin \bigg(\dfrac{x}{3} \bigg)} \, d \bigg(\dfrac{x}{3} \bigg) = -36\cos\bigg(\dfrac{x}{3} \bigg)+ C_{2}$

в)

$\displaystyle \int {(5x + 12) \sin \bigg( \frac{x}{3} \bigg) } \, dx = \int {5x\sin \bigg( \frac{x}{3} \bigg) } \, dx + \int { 12 \sin \bigg( \frac{x}{3} \bigg) } \, dx =$

$\displaystyle = 45\sin\bigg(\dfrac{x}{3} \bigg)-15x\cos\bigg(\dfrac{x}{3} \bigg) + C_{1} -36\cos\bigg(\dfrac{x}{3} \bigg)+ C_{2} =$

$= 45\sin\bigg(\dfrac{x}{3} \bigg)-15x\cos\bigg(\dfrac{x}{3} \bigg) -36\cos\bigg(\dfrac{x}{3} \bigg)+ C$