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Solving integrals/ cos^7xsin^10x

Integral of cos^7xsin^10x dx

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∫01sin⁡10(x)cos⁡7(x) dx\int\limits_{0}^{1} \sin^{10}{\left(x \right)} \cos^{7}{\left(x \right)}\, dx 
Integral(cos(x)^7*sin(x)^10, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    sin⁡10(x)cos⁡7(x)=(1−sin⁡2(x))3sin⁡10(x)cos⁡(x)\sin^{10}{\left(x \right)} \cos^{7}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{3} \sin^{10}{\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:

      ∫(−u16+3u14−3u12+u10) du\int \left(- u^{16} + 3 u^{14} - 3 u^{12} + u^{10}\right)\, du

      1. Integrate term-by-term:

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

          ∫(−u16) du=−∫u16 du\int \left(- u^{16}\right)\, du = - \int u^{16}\, du

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

            ∫u16 du=u1717\int u^{16}\, du = \frac{u^{17}}{17}

          So, the result is: −u1717- \frac{u^{17}}{17}

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

          ∫3u14 du=3∫u14 du\int 3 u^{14}\, du = 3 \int u^{14}\, du

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

            ∫u14 du=u1515\int u^{14}\, du = \frac{u^{15}}{15}

          So, the result is: u155\frac{u^{15}}{5}

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

          ∫(−3u12) du=−3∫u12 du\int \left(- 3 u^{12}\right)\, du = - 3 \int u^{12}\, du

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

            ∫u12 du=u1313\int u^{12}\, du = \frac{u^{13}}{13}

          So, the result is: −3u1313- \frac{3 u^{13}}{13}

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

          ∫u10 du=u1111\int u^{10}\, du = \frac{u^{11}}{11}

        The result is: −u1717+u155−3u1313+u1111- \frac{u^{17}}{17} + \frac{u^{15}}{5} - \frac{3 u^{13}}{13} + \frac{u^{11}}{11}

      Now substitute uu back in:

      −sin⁡17(x)17+sin⁡15(x)5−3sin⁡13(x)13+sin⁡11(x)11- \frac{\sin^{17}{\left(x \right)}}{17} + \frac{\sin^{15}{\left(x \right)}}{5} - \frac{3 \sin^{13}{\left(x \right)}}{13} + \frac{\sin^{11}{\left(x \right)}}{11}

    Method #2

    1. Rewrite the integrand:

      (1−sin⁡2(x))3sin⁡10(x)cos⁡(x)=−sin⁡16(x)cos⁡(x)+3sin⁡14(x)cos⁡(x)−3sin⁡12(x)cos⁡(x)+sin⁡10(x)cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right)^{3} \sin^{10}{\left(x \right)} \cos{\left(x \right)} = - \sin^{16}{\left(x \right)} \cos{\left(x \right)} + 3 \sin^{14}{\left(x \right)} \cos{\left(x \right)} - 3 \sin^{12}{\left(x \right)} \cos{\left(x \right)} + \sin^{10}{\left(x \right)} \cos{\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⁡16(x)cos⁡(x)) dx=−∫sin⁡16(x)cos⁡(x) dx\int \left(- \sin^{16}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{16}{\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:

          ∫u16 du\int u^{16}\, du

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

            ∫u16 du=u1717\int u^{16}\, du = \frac{u^{17}}{17}

          Now substitute uu back in:

          sin⁡17(x)17\frac{\sin^{17}{\left(x \right)}}{17}

        So, the result is: −sin⁡17(x)17- \frac{\sin^{17}{\left(x \right)}}{17}

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

        ∫3sin⁡14(x)cos⁡(x) dx=3∫sin⁡14(x)cos⁡(x) dx\int 3 \sin^{14}{\left(x \right)} \cos{\left(x \right)}\, dx = 3 \int \sin^{14}{\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:

          ∫u14 du\int u^{14}\, du

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

            ∫u14 du=u1515\int u^{14}\, du = \frac{u^{15}}{15}

          Now substitute uu back in:

          sin⁡15(x)15\frac{\sin^{15}{\left(x \right)}}{15}

        So, the result is: sin⁡15(x)5\frac{\sin^{15}{\left(x \right)}}{5}

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

        ∫(−3sin⁡12(x)cos⁡(x)) dx=−3∫sin⁡12(x)cos⁡(x) dx\int \left(- 3 \sin^{12}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 3 \int \sin^{12}{\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:

          ∫u12 du\int u^{12}\, du

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

            ∫u12 du=u1313\int u^{12}\, du = \frac{u^{13}}{13}

          Now substitute uu back in:

          sin⁡13(x)13\frac{\sin^{13}{\left(x \right)}}{13}

        So, the result is: −3sin⁡13(x)13- \frac{3 \sin^{13}{\left(x \right)}}{13}

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

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

        ∫u10 du\int u^{10}\, du

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

          ∫u10 du=u1111\int u^{10}\, du = \frac{u^{11}}{11}

        Now substitute uu back in:

        sin⁡11(x)11\frac{\sin^{11}{\left(x \right)}}{11}

      The result is: −sin⁡17(x)17+sin⁡15(x)5−3sin⁡13(x)13+sin⁡11(x)11- \frac{\sin^{17}{\left(x \right)}}{17} + \frac{\sin^{15}{\left(x \right)}}{5} - \frac{3 \sin^{13}{\left(x \right)}}{13} + \frac{\sin^{11}{\left(x \right)}}{11}

    Method #3

    1. Rewrite the integrand:

      (1−sin⁡2(x))3sin⁡10(x)cos⁡(x)=−sin⁡16(x)cos⁡(x)+3sin⁡14(x)cos⁡(x)−3sin⁡12(x)cos⁡(x)+sin⁡10(x)cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right)^{3} \sin^{10}{\left(x \right)} \cos{\left(x \right)} = - \sin^{16}{\left(x \right)} \cos{\left(x \right)} + 3 \sin^{14}{\left(x \right)} \cos{\left(x \right)} - 3 \sin^{12}{\left(x \right)} \cos{\left(x \right)} + \sin^{10}{\left(x \right)} \cos{\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⁡16(x)cos⁡(x)) dx=−∫sin⁡16(x)cos⁡(x) dx\int \left(- \sin^{16}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{16}{\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:

          ∫u16 du\int u^{16}\, du

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

            ∫u16 du=u1717\int u^{16}\, du = \frac{u^{17}}{17}

          Now substitute uu back in:

          sin⁡17(x)17\frac{\sin^{17}{\left(x \right)}}{17}

        So, the result is: −sin⁡17(x)17- \frac{\sin^{17}{\left(x \right)}}{17}

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

        ∫3sin⁡14(x)cos⁡(x) dx=3∫sin⁡14(x)cos⁡(x) dx\int 3 \sin^{14}{\left(x \right)} \cos{\left(x \right)}\, dx = 3 \int \sin^{14}{\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:

          ∫u14 du\int u^{14}\, du

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

            ∫u14 du=u1515\int u^{14}\, du = \frac{u^{15}}{15}

          Now substitute uu back in:

          sin⁡15(x)15\frac{\sin^{15}{\left(x \right)}}{15}

        So, the result is: sin⁡15(x)5\frac{\sin^{15}{\left(x \right)}}{5}

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

        ∫(−3sin⁡12(x)cos⁡(x)) dx=−3∫sin⁡12(x)cos⁡(x) dx\int \left(- 3 \sin^{12}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 3 \int \sin^{12}{\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:

          ∫u12 du\int u^{12}\, du

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

            ∫u12 du=u1313\int u^{12}\, du = \frac{u^{13}}{13}

          Now substitute uu back in:

          sin⁡13(x)13\frac{\sin^{13}{\left(x \right)}}{13}

        So, the result is: −3sin⁡13(x)13- \frac{3 \sin^{13}{\left(x \right)}}{13}

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

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

        ∫u10 du\int u^{10}\, du

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

          ∫u10 du=u1111\int u^{10}\, du = \frac{u^{11}}{11}

        Now substitute uu back in:

        sin⁡11(x)11\frac{\sin^{11}{\left(x \right)}}{11}

      The result is: −sin⁡17(x)17+sin⁡15(x)5−3sin⁡13(x)13+sin⁡11(x)11- \frac{\sin^{17}{\left(x \right)}}{17} + \frac{\sin^{15}{\left(x \right)}}{5} - \frac{3 \sin^{13}{\left(x \right)}}{13} + \frac{\sin^{11}{\left(x \right)}}{11}

  3. Now simplify:

    (−715sin⁡6(x)+2431sin⁡4(x)−2805sin⁡2(x)+1105)sin⁡11(x)12155\frac{\left(- 715 \sin^{6}{\left(x \right)} + 2431 \sin^{4}{\left(x \right)} - 2805 \sin^{2}{\left(x \right)} + 1105\right) \sin^{11}{\left(x \right)}}{12155}

  4. Add the constant of integration:

    (−715sin⁡6(x)+2431sin⁡4(x)−2805sin⁡2(x)+1105)sin⁡11(x)12155+constant\frac{\left(- 715 \sin^{6}{\left(x \right)} + 2431 \sin^{4}{\left(x \right)} - 2805 \sin^{2}{\left(x \right)} + 1105\right) \sin^{11}{\left(x \right)}}{12155}+ \mathrm{constant}

The answer is:

(−715sin⁡6(x)+2431sin⁡4(x)−2805sin⁡2(x)+1105)sin⁡11(x)12155+constant\frac{\left(- 715 \sin^{6}{\left(x \right)} + 2431 \sin^{4}{\left(x \right)} - 2805 \sin^{2}{\left(x \right)} + 1105\right) \sin^{11}{\left(x \right)}}{12155}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                     
 |                                13         17         15         11   
 |    7       10             3*sin  (x)   sin  (x)   sin  (x)   sin  (x)
 | cos (x)*sin  (x) dx = C - ---------- - -------- + -------- + --------
 |                               13          17         5          11   
/
∫sin⁡10(x)cos⁡7(x) dx=C−sin⁡17(x)17+sin⁡15(x)5−3sin⁡13(x)13+sin⁡11(x)11\int \sin^{10}{\left(x \right)} \cos^{7}{\left(x \right)}\, dx = C - \frac{\sin^{17}{\left(x \right)}}{17} + \frac{\sin^{15}{\left(x \right)}}{5} - \frac{3 \sin^{13}{\left(x \right)}}{13} + \frac{\sin^{11}{\left(x \right)}}{11}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.0000.005
Plot the graph f(x)
The answer [src] 
       13         17         15         11   
  3*sin  (1)   sin  (1)   sin  (1)   sin  (1)
- ---------- - -------- + -------- + --------
      13          17         5          11
−3sin⁡13(1)13−sin⁡17(1)17+sin⁡11(1)11+sin⁡15(1)5- \frac{3 \sin^{13}{\left(1 \right)}}{13} - \frac{\sin^{17}{\left(1 \right)}}{17} + \frac{\sin^{11}{\left(1 \right)}}{11} + \frac{\sin^{15}{\left(1 \right)}}{5} 
=
= 
       13         17         15         11   
  3*sin  (1)   sin  (1)   sin  (1)   sin  (1)
- ---------- - -------- + -------- + --------
      13          17         5          11
−3sin⁡13(1)13−sin⁡17(1)17+sin⁡11(1)11+sin⁡15(1)5- \frac{3 \sin^{13}{\left(1 \right)}}{13} - \frac{\sin^{17}{\left(1 \right)}}{17} + \frac{\sin^{11}{\left(1 \right)}}{11} + \frac{\sin^{15}{\left(1 \right)}}{5} 
-3*sin(1)^13/13 - sin(1)^17/17 + sin(1)^15/5 + sin(1)^11/11
Numerical answer [src] 
0.00103319601406097
0.00103319601406097

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

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