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sin^3
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Integral of sin^3 dx

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∫01sin⁡3(x) dx\int\limits_{0}^{1} \sin^{3}{\left(x \right)}\, dx 
Integral(sin(x)^3, (x, 0, 1))
Detail solution

  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)}

  3. Now simplify:

    (cos⁡(2x)−5)cos⁡(x)6\frac{\left(\cos{\left(2 x \right)} - 5\right) \cos{\left(x \right)}}{6}

  4. Add the constant of integration:

    (cos⁡(2x)−5)cos⁡(x)6+constant\frac{\left(\cos{\left(2 x \right)} - 5\right) \cos{\left(x \right)}}{6}+ \mathrm{constant}

The answer is:

(cos⁡(2x)−5)cos⁡(x)6+constant\frac{\left(\cos{\left(2 x \right)} - 5\right) \cos{\left(x \right)}}{6}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                 
 |                              3   
 |    3                      cos (x)
 | sin (x) dx = C - cos(x) + -------
 |                              3   
/
∫sin⁡3(x) dx=C+cos⁡3(x)3−cos⁡(x)\int \sin^{3}{\left(x \right)}\, dx = C + \frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.901-1
Plot the graph f(x)
The answer [src] 
                3   
2            cos (1)
- - cos(1) + -------
3               3
−cos⁡(1)+cos⁡3(1)3+23- \cos{\left(1 \right)} + \frac{\cos^{3}{\left(1 \right)}}{3} + \frac{2}{3} 
=
= 
                3   
2            cos (1)
- - cos(1) + -------
3               3
−cos⁡(1)+cos⁡3(1)3+23- \cos{\left(1 \right)} + \frac{\cos^{3}{\left(1 \right)}}{3} + \frac{2}{3} 
2/3 - cos(1) + cos(1)^3/3
Numerical answer [src] 
0.178940562548858
0.178940562548858
The graph
Integral of sin^3 dx

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

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