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sin(3x-2)dx
Solving integrals/ sin(3x-2)dx

Integral of sin(3x-2)dx dx

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The solution

You have entered [src] 
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∫01sin⁡(3x−2)1 dx\int\limits_{0}^{1} \sin{\left(3 x - 2 \right)} 1\, dx 
Integral(sin(3*x - 1*2)*1, (x, 0, 1))
Detail solution

  1. Let u=3x−2u = 3 x - 2.

    Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

    ∫sin⁡(u)9 du\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:

      ∫sin⁡(u)3 du=∫sin⁡(u) du3\int \frac{\sin{\left(u \right)}}{3}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}

      1. The integral of sine is negative cosine:

        ∫sin⁡(u) du=−cos⁡(u)\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}

      So, the result is: −cos⁡(u)3- \frac{\cos{\left(u \right)}}{3}

    Now substitute uu back in:

    −cos⁡(3x−2)3- \frac{\cos{\left(3 x - 2 \right)}}{3}

  2. Now simplify:

    −cos⁡(3x−2)3- \frac{\cos{\left(3 x - 2 \right)}}{3}

  3. Add the constant of integration:

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

The answer is:

−cos⁡(3x−2)3+constant- \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      
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−cos⁡(3 x−2)3-{{\cos \left(3\,x-2\right)}\over{3}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
Plot the graph f(x)
The answer [src] 
  cos(1)   cos(2)
- ------ + ------
    3        3
cos⁡2−cos⁡13{{\cos 2-\cos 1}\over{3}} 
=
= 
  cos(1)   cos(2)
- ------ + ------
    3        3
−cos⁡(1)3+cos⁡(2)3- \frac{\cos{\left(1 \right)}}{3} + \frac{\cos{\left(2 \right)}}{3}
Simplify
Numerical answer [src] 
-0.318816380805094
-0.318816380805094
The graph
Integral of sin(3x-2)dx dx

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

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