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Solving integrals/ sin^10xcos^9x

Integral of sin^10xcos^9x dx

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∫0π2sin⁡10(x)cos⁡9(x) dx\int\limits_{0}^{\frac{\pi}{2}} \sin^{10}{\left(x \right)} \cos^{9}{\left(x \right)}\, dx 
Integral(sin(x)^10*cos(x)^9, (x, 0, pi/2))
Detail solution

  1. Rewrite the integrand:

    sin⁡10(x)cos⁡9(x)=(1−sin⁡2(x))4sin⁡10(x)cos⁡(x)\sin^{10}{\left(x \right)} \cos^{9}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{4} \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:

      ∫(u18−4u16+6u14−4u12+u10) du\int \left(u^{18} - 4 u^{16} + 6 u^{14} - 4 u^{12} + u^{10}\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:

          ∫u18 du=u1919\int u^{18}\, du = \frac{u^{19}}{19}

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

          ∫(−4u16) du=−4∫u16 du\int \left(- 4 u^{16}\right)\, du = - 4 \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: −4u1717- \frac{4 u^{17}}{17}

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

          ∫6u14 du=6∫u14 du\int 6 u^{14}\, du = 6 \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: 2u155\frac{2 u^{15}}{5}

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

          ∫(−4u12) du=−4∫u12 du\int \left(- 4 u^{12}\right)\, du = - 4 \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: −4u1313- \frac{4 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: u1919−4u1717+2u155−4u1313+u1111\frac{u^{19}}{19} - \frac{4 u^{17}}{17} + \frac{2 u^{15}}{5} - \frac{4 u^{13}}{13} + \frac{u^{11}}{11}

      Now substitute uu back in:

      sin⁡19(x)19−4sin⁡17(x)17+2sin⁡15(x)5−4sin⁡13(x)13+sin⁡11(x)11\frac{\sin^{19}{\left(x \right)}}{19} - \frac{4 \sin^{17}{\left(x \right)}}{17} + \frac{2 \sin^{15}{\left(x \right)}}{5} - \frac{4 \sin^{13}{\left(x \right)}}{13} + \frac{\sin^{11}{\left(x \right)}}{11}

    Method #2

    1. Rewrite the integrand:

      (1−sin⁡2(x))4sin⁡10(x)cos⁡(x)=sin⁡18(x)cos⁡(x)−4sin⁡16(x)cos⁡(x)+6sin⁡14(x)cos⁡(x)−4sin⁡12(x)cos⁡(x)+sin⁡10(x)cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right)^{4} \sin^{10}{\left(x \right)} \cos{\left(x \right)} = \sin^{18}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{16}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{14}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{12}{\left(x \right)} \cos{\left(x \right)} + \sin^{10}{\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:

        ∫u18 du\int u^{18}\, du

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

          ∫u18 du=u1919\int u^{18}\, du = \frac{u^{19}}{19}

        Now substitute uu back in:

        sin⁡19(x)19\frac{\sin^{19}{\left(x \right)}}{19}

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

        ∫(−4sin⁡16(x)cos⁡(x)) dx=−4∫sin⁡16(x)cos⁡(x) dx\int \left(- 4 \sin^{16}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \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: −4sin⁡17(x)17- \frac{4 \sin^{17}{\left(x \right)}}{17}

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

        ∫6sin⁡14(x)cos⁡(x) dx=6∫sin⁡14(x)cos⁡(x) dx\int 6 \sin^{14}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \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: 2sin⁡15(x)5\frac{2 \sin^{15}{\left(x \right)}}{5}

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

        ∫(−4sin⁡12(x)cos⁡(x)) dx=−4∫sin⁡12(x)cos⁡(x) dx\int \left(- 4 \sin^{12}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \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: −4sin⁡13(x)13- \frac{4 \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⁡19(x)19−4sin⁡17(x)17+2sin⁡15(x)5−4sin⁡13(x)13+sin⁡11(x)11\frac{\sin^{19}{\left(x \right)}}{19} - \frac{4 \sin^{17}{\left(x \right)}}{17} + \frac{2 \sin^{15}{\left(x \right)}}{5} - \frac{4 \sin^{13}{\left(x \right)}}{13} + \frac{\sin^{11}{\left(x \right)}}{11}

    Method #3

    1. Rewrite the integrand:

      (1−sin⁡2(x))4sin⁡10(x)cos⁡(x)=sin⁡18(x)cos⁡(x)−4sin⁡16(x)cos⁡(x)+6sin⁡14(x)cos⁡(x)−4sin⁡12(x)cos⁡(x)+sin⁡10(x)cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right)^{4} \sin^{10}{\left(x \right)} \cos{\left(x \right)} = \sin^{18}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{16}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{14}{\left(x \right)} \cos{\left(x \right)} - 4 \sin^{12}{\left(x \right)} \cos{\left(x \right)} + \sin^{10}{\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:

        ∫u18 du\int u^{18}\, du

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

          ∫u18 du=u1919\int u^{18}\, du = \frac{u^{19}}{19}

        Now substitute uu back in:

        sin⁡19(x)19\frac{\sin^{19}{\left(x \right)}}{19}

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

        ∫(−4sin⁡16(x)cos⁡(x)) dx=−4∫sin⁡16(x)cos⁡(x) dx\int \left(- 4 \sin^{16}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \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: −4sin⁡17(x)17- \frac{4 \sin^{17}{\left(x \right)}}{17}

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

        ∫6sin⁡14(x)cos⁡(x) dx=6∫sin⁡14(x)cos⁡(x) dx\int 6 \sin^{14}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \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: 2sin⁡15(x)5\frac{2 \sin^{15}{\left(x \right)}}{5}

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

        ∫(−4sin⁡12(x)cos⁡(x)) dx=−4∫sin⁡12(x)cos⁡(x) dx\int \left(- 4 \sin^{12}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 4 \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: −4sin⁡13(x)13- \frac{4 \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⁡19(x)19−4sin⁡17(x)17+2sin⁡15(x)5−4sin⁡13(x)13+sin⁡11(x)11\frac{\sin^{19}{\left(x \right)}}{19} - \frac{4 \sin^{17}{\left(x \right)}}{17} + \frac{2 \sin^{15}{\left(x \right)}}{5} - \frac{4 \sin^{13}{\left(x \right)}}{13} + \frac{\sin^{11}{\left(x \right)}}{11}

  3. Now simplify:

    (12155sin⁡8(x)−54340sin⁡6(x)+92378sin⁡4(x)−71060sin⁡2(x)+20995)sin⁡11(x)230945\frac{\left(12155 \sin^{8}{\left(x \right)} - 54340 \sin^{6}{\left(x \right)} + 92378 \sin^{4}{\left(x \right)} - 71060 \sin^{2}{\left(x \right)} + 20995\right) \sin^{11}{\left(x \right)}}{230945}

  4. Add the constant of integration:

    (12155sin⁡8(x)−54340sin⁡6(x)+92378sin⁡4(x)−71060sin⁡2(x)+20995)sin⁡11(x)230945+constant\frac{\left(12155 \sin^{8}{\left(x \right)} - 54340 \sin^{6}{\left(x \right)} + 92378 \sin^{4}{\left(x \right)} - 71060 \sin^{2}{\left(x \right)} + 20995\right) \sin^{11}{\left(x \right)}}{230945}+ \mathrm{constant}

The answer is:

(12155sin⁡8(x)−54340sin⁡6(x)+92378sin⁡4(x)−71060sin⁡2(x)+20995)sin⁡11(x)230945+constant\frac{\left(12155 \sin^{8}{\left(x \right)} - 54340 \sin^{6}{\left(x \right)} + 92378 \sin^{4}{\left(x \right)} - 71060 \sin^{2}{\left(x \right)} + 20995\right) \sin^{11}{\left(x \right)}}{230945}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                    
 |                                13           17         11         19           15   
 |    10       9             4*sin  (x)   4*sin  (x)   sin  (x)   sin  (x)   2*sin  (x)
 | sin  (x)*cos (x) dx = C - ---------- - ---------- + -------- + -------- + ----------
 |                               13           17          11         19          5     
/
∫sin⁡10(x)cos⁡9(x) dx=C+sin⁡19(x)19−4sin⁡17(x)17+2sin⁡15(x)5−4sin⁡13(x)13+sin⁡11(x)11\int \sin^{10}{\left(x \right)} \cos^{9}{\left(x \right)}\, dx = C + \frac{\sin^{19}{\left(x \right)}}{19} - \frac{4 \sin^{17}{\left(x \right)}}{17} + \frac{2 \sin^{15}{\left(x \right)}}{5} - \frac{4 \sin^{13}{\left(x \right)}}{13} + \frac{\sin^{11}{\left(x \right)}}{11}
Simplify
The graph
0.00.10.20.30.40.50.60.70.80.91.01.11.21.31.41.50.0000.002
Plot the graph f(x)
The answer [src] 
 128  
------
230945
128230945\frac{128}{230945} 
=
= 
 128  
------
230945
128230945\frac{128}{230945} 
128/230945
Numerical answer [src] 
0.000554244517092814
0.000554244517092814

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

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