Question

Integration of ∫(cos3x+3sinx)dx ​

Answer 1

`\boxed{\pink{\tt I = (1)/(3)sin(3x) - 3cos(x) + C}}`

Explanation:

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

`\implies\displaystyle\sf 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 .
• Now , Rewrite using du and u .

`\implies\displaystyle\sf I = \int cos\ u (1)/(3)du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int (cos\ u)/(3) du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = (1)/(3)\int (cos(u))/(3) + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = (1)/(3) sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = (1)/(3)sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I = (1)/(3)sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I = (1)/(3)sin(3x) - 3cos(x) + C }}}}}`