Integral of sin3x/3 dx

The solution

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  t            
  /            
 |             
 |  sin(3*x)   
 |  -------- dx
 |     3       
 |             
/              
0

\int\limits_{0}^{t} \frac{\sin{\left(3 x \right)}}{3}\, dx

Simplify

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

$\int \frac{\sin{\left(3 x \right)}}{3}\, dx = \frac{\int \sin{\left(3 x \right)}\, dx}{3}$
1. Let $u = 3 x$ .
  
  Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
  
  $\int \frac{\sin{\left(u \right)}}{9}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
    1. The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
    So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
  Now substitute $u$ back in:
  
  $- \frac{\cos{\left(3 x \right)}}{3}$
So, the result is: $- \frac{\cos{\left(3 x \right)}}{9}$
Add the constant of integration:

$- \frac{\cos{\left(3 x \right)}}{9}+ \mathrm{constant}$

The answer is:

- \frac{\cos{\left(3 x \right)}}{9}+ \mathrm{constant}

The answer (Indefinite) [src]

  /                          
 |                           
 | sin(3*x)          cos(3*x)
 | -------- dx = C - --------
 |    3                 9    
 |                           
/

-{{\cos \left(3\,x\right)}\over{9}}

Simplify

The answer [src]

1   cos(3*t)
- - --------
9      9

- \frac{\cos{\left(3 t \right)}}{9} + \frac{1}{9}

1   cos(3*t)
- - --------
9      9

- \frac{\cos{\left(3 t \right)}}{9} + \frac{1}{9}

Simplify

Use the examples entering the upper and lower limits of integration.

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Integral of sin3x/3 dx

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