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

Integral of 2sin(3x)-3x^2 dx

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The solution

You have entered [src] 
  2                       
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 |  \2*sin(3*x) - 3*x / dx
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∫02(−3x2+2sin⁡(3x)) dx\int\limits_{0}^{2} \left(- 3 x^{2} + 2 \sin{\left(3 x \right)}\right)\, dx 
Integral(2*sin(3*x) - 3*x^2, (x, 0, 2))
Detail solution

  1. Integrate term-by-term:

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

      ∫(−3x2) dx=−3∫x2 dx\int \left(- 3 x^{2}\right)\, dx = - 3 \int x^{2}\, dx

      1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫x2 dx=x33\int x^{2}\, dx = \frac{x^{3}}{3}

      So, the result is: −x3- x^{3}

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

      ∫2sin⁡(3x) dx=2∫sin⁡(3x) dx\int 2 \sin{\left(3 x \right)}\, dx = 2 \int \sin{\left(3 x \right)}\, dx

      1. Let u=3xu = 3 x.

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

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

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

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

      So, the result is: −2cos⁡(3x)3- \frac{2 \cos{\left(3 x \right)}}{3}

    The result is: −x3−2cos⁡(3x)3- x^{3} - \frac{2 \cos{\left(3 x \right)}}{3}

  2. Add the constant of integration:

    −x3−2cos⁡(3x)3+constant- x^{3} - \frac{2 \cos{\left(3 x \right)}}{3}+ \mathrm{constant}

The answer is:

−x3−2cos⁡(3x)3+constant- x^{3} - \frac{2 \cos{\left(3 x \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                            
 |                                             
 | /                2\           3   2*cos(3*x)
 | \2*sin(3*x) - 3*x / dx = C - x  - ----------
 |                                       3     
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∫(−3x2+2sin⁡(3x)) dx=C−x3−2cos⁡(3x)3\int \left(- 3 x^{2} + 2 \sin{\left(3 x \right)}\right)\, dx = C - x^{3} - \frac{2 \cos{\left(3 x \right)}}{3}
Simplify
The graph
0.02.00.20.40.60.81.01.21.41.61.8-2020
Plot the graph f(x)
The answer [src] 
  22   2*cos(6)
- -- - --------
  3       3
−223−2cos⁡(6)3- \frac{22}{3} - \frac{2 \cos{\left(6 \right)}}{3} 
=
= 
  22   2*cos(6)
- -- - --------
  3       3
−223−2cos⁡(6)3- \frac{22}{3} - \frac{2 \cos{\left(6 \right)}}{3} 
-22/3 - 2*cos(6)/3
Numerical answer [src] 
-7.97344685776691
-7.97344685776691

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

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