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7 votes
7 votes
∫(cos3x+3sinx)dx integration​

User Rabin
by
2.7k points

2 Answers

20 votes
20 votes
(47)83-x (cos7y) + cos34
User Lmop
by
2.8k points
18 votes
18 votes

Answer:


I = (1)/(3)sin(3x) - 3cos(x) + C

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

User Nitish Patel
by
2.9k points
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