Mathematics: Can someone help me solve this?

Step-by-step explanation:We need to integrate the given expression. Let I be the answer .Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .Rewrite using du and u ....

Can someone help me solve this? - id14799097, mustafajibawi1

Ответ:

AeelynRamos

25.11.2021

Mathematics

$\\ \sf\longmapsto 3q+4+9=-14$

$\\ \sf\longmapsto 3q+13=-14$

$\\ \sf\longmapsto 3q=-14-13$

$\\ \sf\longmapsto 3q=-27$

$\\ \sf\longmapsto q=-9$

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Ответ:

tonta22

23.06.2020

Mathematics

$I = \dfrac{1}{3}sin(3x) - 3cos(x) + C$

Step-by-step explanation:

We need to integrate the given expression. Let I be the answer .

$\implies\displaystyle I = \int (cos(3x) + 3sin(x) )dx \\\\\implies\displaystyle I = \int cos(3x) + \int sin(x)\ dx$

Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .Rewrite using du and u .

$\implies\displaystyle I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle I = \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \boxed{\boxed{\displaystyle I = \dfrac{1}{3}sin(3x) - 3cos(x) + C }}$

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