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sin^2xcos^5xdx
Solving integrals/ sin^2xcos^5xdx

Integral of sin^2xcos^5xdx dx

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∫01sin⁡2(x)cos⁡5(x)1 dx\int\limits_{0}^{1} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} 1\, dx 
Integral(sin(x)^2*cos(x)^5*1, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    sin⁡2(x)cos⁡5(x)1=(1−sin⁡2(x))2sin⁡2(x)cos⁡(x)\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} 1 = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{2}{\left(x \right)} \cos{\left(x \right)}

  2. There are multiple ways to do this integral.

    Method #1

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

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

      ∫(u6−2u4+u2) du\int \left(u^{6} - 2 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:

          ∫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:

          ∫(−2u4) du=−2∫u4 du\int \left(- 2 u^{4}\right)\, du = - 2 \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: −2u55- \frac{2 u^{5}}{5}

        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}

        The result is: u77−2u55+u33\frac{u^{7}}{7} - \frac{2 u^{5}}{5} + \frac{u^{3}}{3}

      Now substitute uu back in:

      sin⁡7(x)7−2sin⁡5(x)5+sin⁡3(x)3\frac{\sin^{7}{\left(x \right)}}{7} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}

    Method #2

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

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

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

        ∫u6 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}

        Now substitute uu back in:

        sin⁡7(x)7\frac{\sin^{7}{\left(x \right)}}{7}

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

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

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

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

          ∫u4 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}

          Now substitute uu back in:

          sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

        So, the result is: −2sin⁡5(x)5- \frac{2 \sin^{5}{\left(x \right)}}{5}

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

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

        ∫u2 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}

        Now substitute uu back in:

        sin⁡3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

      The result is: sin⁡7(x)7−2sin⁡5(x)5+sin⁡3(x)3\frac{\sin^{7}{\left(x \right)}}{7} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}

    Method #3

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

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

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

        ∫u6 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}

        Now substitute uu back in:

        sin⁡7(x)7\frac{\sin^{7}{\left(x \right)}}{7}

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

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

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

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

          ∫u4 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}

          Now substitute uu back in:

          sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

        So, the result is: −2sin⁡5(x)5- \frac{2 \sin^{5}{\left(x \right)}}{5}

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

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

        ∫u2 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}

        Now substitute uu back in:

        sin⁡3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

      The result is: sin⁡7(x)7−2sin⁡5(x)5+sin⁡3(x)3\frac{\sin^{7}{\left(x \right)}}{7} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}

  3. Now simplify:

    (15sin⁡4(x)−42sin⁡2(x)+35)sin⁡3(x)105\frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105}

  4. Add the constant of integration:

    (15sin⁡4(x)−42sin⁡2(x)+35)sin⁡3(x)105+constant\frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105}+ \mathrm{constant}

The answer is:

(15sin⁡4(x)−42sin⁡2(x)+35)sin⁡3(x)105+constant\frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                        
 |                                 5         3         7   
 |    2       5               2*sin (x)   sin (x)   sin (x)
 | sin (x)*cos (x)*1 dx = C - --------- + ------- + -------
 |                                5          3         7   
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15 sin⁡7x−42 sin⁡5x+35 sin⁡3x105{{15\,\sin ^7x-42\,\sin ^5x+35\,\sin ^3x}\over{105}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.00.2
Plot the graph f(x)
The answer [src] 
       5         3         7   
  2*sin (1)   sin (1)   sin (1)
- --------- + ------- + -------
      5          3         7
15 sin⁡71−42 sin⁡51+35 sin⁡31105{{15\,\sin ^71-42\,\sin ^51+35\,\sin ^31}\over{105}} 
=
= 
       5         3         7   
  2*sin (1)   sin (1)   sin (1)
- --------- + ------- + -------
      5          3         7
−2sin⁡5(1)5+sin⁡7(1)7+sin⁡3(1)3- \frac{2 \sin^{5}{\left(1 \right)}}{5} + \frac{\sin^{7}{\left(1 \right)}}{7} + \frac{\sin^{3}{\left(1 \right)}}{3}
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Numerical answer [src] 
0.0725283477775366
0.0725283477775366
The graph
Integral of sin^2xcos^5xdx dx

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

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