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Solving integrals/ sinx-sin^3x/3

Integral of sinx-sin^3x/3 dx

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∫01(−sin⁡3(x)3+sin⁡(x)) dx\int\limits_{0}^{1} \left(- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right)\, dx 
Integral(sin(x) - sin(x)^3/3, (x, 0, 1))
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:

      ∫(−sin⁡3(x)3) dx=−∫sin⁡3(x) dx3\int \left(- \frac{\sin^{3}{\left(x \right)}}{3}\right)\, dx = - \frac{\int \sin^{3}{\left(x \right)}\, dx}{3}

      1. Rewrite the integrand:

        sin⁡3(x)=(1−cos⁡2(x))sin⁡(x)\sin^{3}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}

      2. There are multiple ways to do this integral.

        Method #1

        1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

          Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute dudu:

          ∫(u2−1) du\int \left(u^{2} - 1\right)\, du

          1. Integrate term-by-term:

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

              ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

            1. The integral of a constant is the constant times the variable of integration:

              ∫(−1) du=−u\int \left(-1\right)\, du = - u

            The result is: u33−u\frac{u^{3}}{3} - u

          Now substitute uu back in:

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

        Method #2

        1. Rewrite the integrand:

          (1−cos⁡2(x))sin⁡(x)=−sin⁡(x)cos⁡2(x)+sin⁡(x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} = - \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}

        2. Integrate term-by-term:

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

            ∫(−sin⁡(x)cos⁡2(x)) dx=−∫sin⁡(x)cos⁡2(x) dx\int \left(- \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

            1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

              Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

              ∫(−u2) du\int \left(- u^{2}\right)\, du

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

                ∫u2 du=−∫u2 du\int u^{2}\, du = - \int u^{2}\, du

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                  ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

                So, the result is: −u33- \frac{u^{3}}{3}

              Now substitute uu back in:

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

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

          1. The integral of sine is negative cosine:

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

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

        Method #3

        1. Rewrite the integrand:

          (1−cos⁡2(x))sin⁡(x)=−sin⁡(x)cos⁡2(x)+sin⁡(x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} = - \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}

        2. Integrate term-by-term:

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

            ∫(−sin⁡(x)cos⁡2(x)) dx=−∫sin⁡(x)cos⁡2(x) dx\int \left(- \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

            1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

              Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

              ∫(−u2) du\int \left(- u^{2}\right)\, du

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

                ∫u2 du=−∫u2 du\int u^{2}\, du = - \int u^{2}\, du

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                  ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

                So, the result is: −u33- \frac{u^{3}}{3}

              Now substitute uu back in:

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

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

          1. The integral of sine is negative cosine:

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

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

      So, the result is: −cos⁡3(x)9+cos⁡(x)3- \frac{\cos^{3}{\left(x \right)}}{9} + \frac{\cos{\left(x \right)}}{3}

    1. The integral of sine is negative cosine:

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

    The result is: −cos⁡3(x)9−2cos⁡(x)3- \frac{\cos^{3}{\left(x \right)}}{9} - \frac{2 \cos{\left(x \right)}}{3}

  2. Now simplify:

    −(cos⁡2(x)+6)cos⁡(x)9- \frac{\left(\cos^{2}{\left(x \right)} + 6\right) \cos{\left(x \right)}}{9}

  3. Add the constant of integration:

    −(cos⁡2(x)+6)cos⁡(x)9+constant- \frac{\left(\cos^{2}{\left(x \right)} + 6\right) \cos{\left(x \right)}}{9}+ \mathrm{constant}

The answer is:

−(cos⁡2(x)+6)cos⁡(x)9+constant- \frac{\left(\cos^{2}{\left(x \right)} + 6\right) \cos{\left(x \right)}}{9}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                              
 |                                               
 | /            3   \                        3   
 | |         sin (x)|          2*cos(x)   cos (x)
 | |sin(x) - -------| dx = C - -------- - -------
 | \            3   /             3          9   
 |                                               
/
∫(−sin⁡3(x)3+sin⁡(x)) dx=C−cos⁡3(x)9−2cos⁡(x)3\int \left(- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}\right)\, dx = C - \frac{\cos^{3}{\left(x \right)}}{9} - \frac{2 \cos{\left(x \right)}}{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] 
                  3   
7   2*cos(1)   cos (1)
- - -------- - -------
9      3          9
−2cos⁡(1)3−cos⁡3(1)9+79- \frac{2 \cos{\left(1 \right)}}{3} - \frac{\cos^{3}{\left(1 \right)}}{9} + \frac{7}{9} 
=
= 
                  3   
7   2*cos(1)   cos (1)
- - -------- - -------
9      3          9
−2cos⁡(1)3−cos⁡3(1)9+79- \frac{2 \cos{\left(1 \right)}}{3} - \frac{\cos^{3}{\left(1 \right)}}{9} + \frac{7}{9} 
7/9 - 2*cos(1)/3 - cos(1)^3/9
Numerical answer [src] 
0.400050839948908
0.400050839948908

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

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