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sin³xcos²x
Solving integrals/ sin³xcos²x

Integral of sin³xcos²x dx

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∫01sin⁡3(x)cos⁡2(x) dx\int\limits_{0}^{1} \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx 
Integral(sin(x)^3*cos(x)^2, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

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

      ∫(u4−u2) du\int \left(u^{4} - u^{2}\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:

          ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

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

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

        The result is: u55−u33\frac{u^{5}}{5} - \frac{u^{3}}{3}

      Now substitute uu back in:

      cos⁡5(x)5−cos⁡3(x)3\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

    Method #2

    1. Rewrite the integrand:

      (1−cos⁡2(x))sin⁡(x)cos⁡2(x)=−sin⁡(x)cos⁡4(x)+sin⁡(x)cos⁡2(x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} = - \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\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⁡4(x)) dx=−∫sin⁡(x)cos⁡4(x) dx\int \left(- \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{4}{\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:

          ∫u4 du\int u^{4}\, du

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

            ∫(−u4) du=−∫u4 du\int \left(- u^{4}\right)\, du = - \int u^{4}\, du

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

              ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

            So, the result is: −u55- \frac{u^{5}}{5}

          Now substitute uu back in:

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

        So, the result is: cos⁡5(x)5\frac{\cos^{5}{\left(x \right)}}{5}

      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 u^{2}\, du

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

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

      The result is: cos⁡5(x)5−cos⁡3(x)3\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

    Method #3

    1. Rewrite the integrand:

      (1−cos⁡2(x))sin⁡(x)cos⁡2(x)=−sin⁡(x)cos⁡4(x)+sin⁡(x)cos⁡2(x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} = - \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\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⁡4(x)) dx=−∫sin⁡(x)cos⁡4(x) dx\int \left(- \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{4}{\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:

          ∫u4 du\int u^{4}\, du

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

            ∫(−u4) du=−∫u4 du\int \left(- u^{4}\right)\, du = - \int u^{4}\, du

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

              ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

            So, the result is: −u55- \frac{u^{5}}{5}

          Now substitute uu back in:

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

        So, the result is: cos⁡5(x)5\frac{\cos^{5}{\left(x \right)}}{5}

      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 u^{2}\, du

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

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

      The result is: cos⁡5(x)5−cos⁡3(x)3\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}

  3. Add the constant of integration:

    cos⁡5(x)5−cos⁡3(x)3+constant\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}+ \mathrm{constant}

The answer is:

cos⁡5(x)5−cos⁡3(x)3+constant\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                          
 |                             3         5   
 |    3       2             cos (x)   cos (x)
 | sin (x)*cos (x) dx = C - ------- + -------
 |                             3         5   
/
3 cos⁡5x−5 cos⁡3x15{{3\,\cos ^5x-5\,\cos ^3x}\over{15}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.5-0.5
Plot the graph f(x)
The answer [src] 
        3         5   
2    cos (1)   cos (1)
-- - ------- + -------
15      3         5
3 cos⁡51−5 cos⁡3115+215{{3\,\cos ^51-5\,\cos ^31}\over{15}}+{{2}\over{15}} 
=
= 
        3         5   
2    cos (1)   cos (1)
-- - ------- + -------
15      3         5
−cos⁡3(1)3+cos⁡5(1)5+215- \frac{\cos^{3}{\left(1 \right)}}{3} + \frac{\cos^{5}{\left(1 \right)}}{5} + \frac{2}{15}
Simplify
Numerical answer [src] 
0.0899661660972821
0.0899661660972821
The graph
Integral of sin³xcos²x dx

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

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