Question

Integration of ∫(cos3x+3sinx)dx ​

Answer 1

Answer:

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

Step-by-step explanation:

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

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

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

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