Question

Integration of ∫(cos3x+3sinx)dx integration​

Answer 1

Answer:

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

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 .
• Rewrite using du and u .

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

Answer 2

Answer:

$\displaystyle \large{\frac{\sin 3x}{3} - 3\cos x + C}$

Step-by-step explanation:

We are given the indefinite integral:—

$\displaystyle \large{\int (\cos 3x + 3\sin x) \ dx}$

Important Formulas

$\displaystyle \large{\int f(ax+b) \ dx = \frac{1}{a} F(ax+b) + C}\\\displaystyle \large{\int \cos(ax) \ dx = \frac{1}{a} \sin (ax) + C \ \ \tt{(a \ \ is \ \ a \ \ constant.)}}\\\displaystyle \large{\int \sin x \ dx = - \cos x + C}\\\displaystyle \large{\int [f(x) \pm g(x)] \ dx = \int f(x) \ dx \pm \int g(x) \ dx}\\\displaystyle \large{\int kf(x) \ dx = k \int f(x) \ dx \ \ (\tt{k \ \ is \ \ a \ \ constant.})}$

Therefore, from the integral, apply the properties above:—

$\displaystyle \large{\int (\cos 3x + 3\sin x) \ dx = \int \cos 3x \ dx + \int 3 \sin x \ dx}\\\displaystyle \large{\int (\cos 3x + 3\sin x) \ dx = \int \cos 3x \ dx + 3 \int \sin x \ dx}\\\displaystyle \large{\int (\cos 3x + 3\sin x) \ dx = \frac{1}{3} \sin 3x + 3\cdot -\cos x + C}\\\displaystyle \large{\int (\cos 3x + 3\sin x) \ dx = \frac{\sin 3x}{3} - 3\cos x + C}$

Hence, the solution is:—

$\displaystyle \large \boxed{\frac{\sin 3x}{3} - 3\cos x + C}$