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Solving integrals/ sin(3x-2)dx

Integral of sin(3x-2)dx dx

The solution

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  1                  
  /                  
 |                   
 |  sin(3*x - 2)*1 dx
 |                   
/                    
0

$$\int\limits_{0}^{1} \sin{\left(3 x - 2 \right)} 1\, dx$$

Integral(sin(3*x - 1*2)*1, (x, 0, 1))

Detail solution

Let $u = 3 x - 2$ .

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:

$- \frac{\cos{\left(3 x - 2 \right)}}{3}$
Now simplify:

$- \frac{\cos{\left(3 x - 2 \right)}}{3}$
Add the constant of integration:

$- \frac{\cos{\left(3 x - 2 \right)}}{3}+ \mathrm{constant}$

The answer is:

$- \frac{\cos{\left(3 x - 2 \right)}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                    
 |                         cos(3*x - 2)
 | sin(3*x - 2)*1 dx = C - ------------
 |                              3      
/

$$-{{\cos \left(3\,x-2\right)}\over{3}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  cos(1)   cos(2)
- ------ + ------
    3        3

$${{\cos 2-\cos 1}\over{3}}$$

  cos(1)   cos(2)
- ------ + ------
    3        3

$$- \frac{\cos{\left(1 \right)}}{3} + \frac{\cos{\left(2 \right)}}{3}$$

Simplify

Numerical answer [src]

-0.318816380805094

-0.318816380805094

The graph

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

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Integral of sin(3x-2)dx dx

Limits of integration:

The graph:

Piecewise:

The solution