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sin^3xcos^4xdx
Solving integrals/ sin^3xcos^4xdx

Integral of sin^3xcos^4xdx dx

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∫01sin⁡3(x)cos⁡4(x)1 dx\int\limits_{0}^{1} \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} 1\, dx 
Integral(sin(x)^3*cos(x)^4*1, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    sin⁡3(x)cos⁡4(x)1=(1−cos⁡2(x))sin⁡(x)cos⁡4(x)\sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} 1 = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{4}{\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:

      ∫(u6−u4) du\int \left(u^{6} - u^{4}\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:

          ∫u6 du=u77\int u^{6}\, du = \frac{u^{7}}{7}

        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}

        The result is: u77−u55\frac{u^{7}}{7} - \frac{u^{5}}{5}

      Now substitute uu back in:

      cos⁡7(x)7−cos⁡5(x)5\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}

    Method #2

    1. Rewrite the integrand:

      (1−cos⁡2(x))sin⁡(x)cos⁡4(x)=−sin⁡(x)cos⁡6(x)+sin⁡(x)cos⁡4(x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{4}{\left(x \right)} = - \sin{\left(x \right)} \cos^{6}{\left(x \right)} + \sin{\left(x \right)} \cos^{4}{\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⁡6(x)) dx=−∫sin⁡(x)cos⁡6(x) dx\int \left(- \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{6}{\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:

          ∫u6 du\int u^{6}\, du

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

            ∫(−u6) du=−∫u6 du\int \left(- u^{6}\right)\, du = - \int u^{6}\, du

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

              ∫u6 du=u77\int u^{6}\, du = \frac{u^{7}}{7}

            So, the result is: −u77- \frac{u^{7}}{7}

          Now substitute uu back in:

          −cos⁡7(x)7- \frac{\cos^{7}{\left(x \right)}}{7}

        So, the result is: cos⁡7(x)7\frac{\cos^{7}{\left(x \right)}}{7}

      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}

      The result is: cos⁡7(x)7−cos⁡5(x)5\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}

    Method #3

    1. Rewrite the integrand:

      (1−cos⁡2(x))sin⁡(x)cos⁡4(x)=−sin⁡(x)cos⁡6(x)+sin⁡(x)cos⁡4(x)\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{4}{\left(x \right)} = - \sin{\left(x \right)} \cos^{6}{\left(x \right)} + \sin{\left(x \right)} \cos^{4}{\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⁡6(x)) dx=−∫sin⁡(x)cos⁡6(x) dx\int \left(- \sin{\left(x \right)} \cos^{6}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{6}{\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:

          ∫u6 du\int u^{6}\, du

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

            ∫(−u6) du=−∫u6 du\int \left(- u^{6}\right)\, du = - \int u^{6}\, du

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

              ∫u6 du=u77\int u^{6}\, du = \frac{u^{7}}{7}

            So, the result is: −u77- \frac{u^{7}}{7}

          Now substitute uu back in:

          −cos⁡7(x)7- \frac{\cos^{7}{\left(x \right)}}{7}

        So, the result is: cos⁡7(x)7\frac{\cos^{7}{\left(x \right)}}{7}

      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}

      The result is: cos⁡7(x)7−cos⁡5(x)5\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}

  3. Add the constant of integration:

    cos⁡7(x)7−cos⁡5(x)5+constant\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}+ \mathrm{constant}

The answer is:

cos⁡7(x)7−cos⁡5(x)5+constant\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                            
 |                               5         7   
 |    3       4               cos (x)   cos (x)
 | sin (x)*cos (x)*1 dx = C - ------- + -------
 |                               5         7   
/
5 cos⁡7x−7 cos⁡5x35{{5\,\cos ^7x-7\,\cos ^5x}\over{35}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.2-0.2
Plot the graph f(x)
The answer [src] 
        5         7   
2    cos (1)   cos (1)
-- - ------- + -------
35      5         7
5 cos⁡71−7 cos⁡5135+235{{5\,\cos ^71-7\,\cos ^51}\over{35}}+{{2}\over{35}} 
=
= 
        5         7   
2    cos (1)   cos (1)
-- - ------- + -------
35      5         7
−cos⁡5(1)5+cos⁡7(1)7+235- \frac{\cos^{5}{\left(1 \right)}}{5} + \frac{\cos^{7}{\left(1 \right)}}{7} + \frac{2}{35}
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Numerical answer [src] 
0.0498540811762589
0.0498540811762589
The graph
Integral of sin^3xcos^4xdx dx

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

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