Integral of sin9x dx

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  1            
  /            
 |             
 |  sin(9*x) dx
 |             
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0

$$\int\limits_{0}^{1} \sin{\left(9 x \right)}\, dx$$

Integral(sin(9*x), (x, 0, 1))

Detail solution

Let $u = 9 x$ .

Then let $du = 9 dx$ and substitute $\frac{du}{9}$ :

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

$- \frac{\cos{\left(9 x \right)}}{9}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

$$\int \sin{\left(9 x \right)}\, dx = C - \frac{\cos{\left(9 x \right)}}{9}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1   cos(9)
- - ------
9     9

$$\frac{1}{9} - \frac{\cos{\left(9 \right)}}{9}$$

1   cos(9)
- - ------
9     9

$$\frac{1}{9} - \frac{\cos{\left(9 \right)}}{9}$$

1/9 - cos(9)/9

Numerical answer [src]

0.212347806876075

0.212347806876075

The graph

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