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sin^6x*dx
Solving integrals/ sin^6x*dx

Integral of sin^6x*dx dx

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∫01sin⁡6(x)1 dx\int\limits_{0}^{1} \sin^{6}{\left(x \right)} 1\, dx 
Integral(sin(x)^6*1, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    sin⁡6(x)1=(12−cos⁡(2x)2)3\sin^{6}{\left(x \right)} 1 = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{3}

  2. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      (12−cos⁡(2x)2)3=−cos⁡3(2x)8+3cos⁡2(2x)8−3cos⁡(2x)8+18\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{3} = - \frac{\cos^{3}{\left(2 x \right)}}{8} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} - \frac{3 \cos{\left(2 x \right)}}{8} + \frac{1}{8}

    2. Integrate term-by-term:

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

        ∫(−cos⁡3(2x)8) dx=−∫cos⁡3(2x) dx8\int \left(- \frac{\cos^{3}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}

        1. Rewrite the integrand:

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

        2. There are multiple ways to do this integral.

          Method #1

          1. Let u=sin⁡(2x)u = \sin{\left(2 x \right)}.

            Then let du=2cos⁡(2x)dxdu = 2 \cos{\left(2 x \right)} dx and substitute dudu:

            ∫(12−u22) du\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du

            1. Integrate term-by-term:

              1. The integral of a constant is the constant times the variable of integration:

                ∫12 du=u2\int \frac{1}{2}\, du = \frac{u}{2}

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

                ∫(−u22) du=−∫u2 du2\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}

                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: −u36- \frac{u^{3}}{6}

              The result is: −u36+u2- \frac{u^{3}}{6} + \frac{u}{2}

            Now substitute uu back in:

            −sin⁡3(2x)6+sin⁡(2x)2- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}

          Method #2

          1. Rewrite the integrand:

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

              1. Let u=sin⁡(2x)u = \sin{\left(2 x \right)}.

                Then let du=2cos⁡(2x)dxdu = 2 \cos{\left(2 x \right)} dx and substitute du2\frac{du}{2}:

                ∫u24 du\int \frac{u^{2}}{4}\, du

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

                  ∫u22 du=∫u2 du2\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}

                  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: u36\frac{u^{3}}{6}

                Now substitute uu back in:

                sin⁡3(2x)6\frac{\sin^{3}{\left(2 x \right)}}{6}

              So, the result is: −sin⁡3(2x)6- \frac{\sin^{3}{\left(2 x \right)}}{6}

            1. Let u=2xu = 2 x.

              Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

              ∫cos⁡(u)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

                ∫cos⁡(u)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

                So, the result is: sin⁡(u)2\frac{\sin{\left(u \right)}}{2}

              Now substitute uu back in:

              sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

            The result is: −sin⁡3(2x)6+sin⁡(2x)2- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}

          Method #3

          1. Rewrite the integrand:

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

              1. Let u=sin⁡(2x)u = \sin{\left(2 x \right)}.

                Then let du=2cos⁡(2x)dxdu = 2 \cos{\left(2 x \right)} dx and substitute du2\frac{du}{2}:

                ∫u24 du\int \frac{u^{2}}{4}\, du

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

                  ∫u22 du=∫u2 du2\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}

                  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: u36\frac{u^{3}}{6}

                Now substitute uu back in:

                sin⁡3(2x)6\frac{\sin^{3}{\left(2 x \right)}}{6}

              So, the result is: −sin⁡3(2x)6- \frac{\sin^{3}{\left(2 x \right)}}{6}

            1. Let u=2xu = 2 x.

              Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

              ∫cos⁡(u)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

                ∫cos⁡(u)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

                So, the result is: sin⁡(u)2\frac{\sin{\left(u \right)}}{2}

              Now substitute uu back in:

              sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

            The result is: −sin⁡3(2x)6+sin⁡(2x)2- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}

        So, the result is: sin⁡3(2x)48−sin⁡(2x)16\frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{16}

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

        ∫3cos⁡2(2x)8 dx=3∫cos⁡2(2x) dx8\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}

        1. Rewrite the integrand:

          cos⁡2(2x)=cos⁡(4x)2+12\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}

        2. Integrate term-by-term:

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

            ∫cos⁡(4x)2 dx=∫cos⁡(4x) dx2\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}

            1. Let u=4xu = 4 x.

              Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

              ∫cos⁡(u)16 du\int \frac{\cos{\left(u \right)}}{16}\, du

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

                ∫cos⁡(u)4 du=∫cos⁡(u) du4\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

                So, the result is: sin⁡(u)4\frac{\sin{\left(u \right)}}{4}

              Now substitute uu back in:

              sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

            So, the result is: sin⁡(4x)8\frac{\sin{\left(4 x \right)}}{8}

          1. The integral of a constant is the constant times the variable of integration:

            ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

          The result is: x2+sin⁡(4x)8\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}

        So, the result is: 3x16+3sin⁡(4x)64\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}

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

        ∫(−3cos⁡(2x)8) dx=−3∫cos⁡(2x) dx8\int \left(- \frac{3 \cos{\left(2 x \right)}}{8}\right)\, dx = - \frac{3 \int \cos{\left(2 x \right)}\, dx}{8}

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

            ∫cos⁡(u)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

            1. The integral of cosine is sine:

              ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

            So, the result is: sin⁡(u)2\frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

        So, the result is: −3sin⁡(2x)16- \frac{3 \sin{\left(2 x \right)}}{16}

      1. The integral of a constant is the constant times the variable of integration:

        ∫18 dx=x8\int \frac{1}{8}\, dx = \frac{x}{8}

      The result is: 5x16+sin⁡3(2x)48−sin⁡(2x)4+3sin⁡(4x)64\frac{5 x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}

    Method #2

    1. Rewrite the integrand:

      (12−cos⁡(2x)2)3=−cos⁡3(2x)8+3cos⁡2(2x)8−3cos⁡(2x)8+18\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{3} = - \frac{\cos^{3}{\left(2 x \right)}}{8} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} - \frac{3 \cos{\left(2 x \right)}}{8} + \frac{1}{8}

    2. Integrate term-by-term:

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

        ∫(−cos⁡3(2x)8) dx=−∫cos⁡3(2x) dx8\int \left(- \frac{\cos^{3}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}

        1. Rewrite the integrand:

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

        2. Let u=sin⁡(2x)u = \sin{\left(2 x \right)}.

          Then let du=2cos⁡(2x)dxdu = 2 \cos{\left(2 x \right)} dx and substitute dudu:

          ∫(12−u22) du\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du

          1. Integrate term-by-term:

            1. The integral of a constant is the constant times the variable of integration:

              ∫12 du=u2\int \frac{1}{2}\, du = \frac{u}{2}

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

              ∫(−u22) du=−∫u2 du2\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}

              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: −u36- \frac{u^{3}}{6}

            The result is: −u36+u2- \frac{u^{3}}{6} + \frac{u}{2}

          Now substitute uu back in:

          −sin⁡3(2x)6+sin⁡(2x)2- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}

        So, the result is: sin⁡3(2x)48−sin⁡(2x)16\frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{16}

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

        ∫3cos⁡2(2x)8 dx=3∫cos⁡2(2x) dx8\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}

        1. Rewrite the integrand:

          cos⁡2(2x)=cos⁡(4x)2+12\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}

        2. Integrate term-by-term:

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

            ∫cos⁡(4x)2 dx=∫cos⁡(4x) dx2\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}

            1. Let u=4xu = 4 x.

              Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

              ∫cos⁡(u)16 du\int \frac{\cos{\left(u \right)}}{16}\, du

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

                ∫cos⁡(u)4 du=∫cos⁡(u) du4\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

                1. The integral of cosine is sine:

                  ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

                So, the result is: sin⁡(u)4\frac{\sin{\left(u \right)}}{4}

              Now substitute uu back in:

              sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

            So, the result is: sin⁡(4x)8\frac{\sin{\left(4 x \right)}}{8}

          1. The integral of a constant is the constant times the variable of integration:

            ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

          The result is: x2+sin⁡(4x)8\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}

        So, the result is: 3x16+3sin⁡(4x)64\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}

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

        ∫(−3cos⁡(2x)8) dx=−3∫cos⁡(2x) dx8\int \left(- \frac{3 \cos{\left(2 x \right)}}{8}\right)\, dx = - \frac{3 \int \cos{\left(2 x \right)}\, dx}{8}

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫cos⁡(u)4 du\int \frac{\cos{\left(u \right)}}{4}\, du

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

            ∫cos⁡(u)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

            1. The integral of cosine is sine:

              ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

            So, the result is: sin⁡(u)2\frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

        So, the result is: −3sin⁡(2x)16- \frac{3 \sin{\left(2 x \right)}}{16}

      1. The integral of a constant is the constant times the variable of integration:

        ∫18 dx=x8\int \frac{1}{8}\, dx = \frac{x}{8}

      The result is: 5x16+sin⁡3(2x)48−sin⁡(2x)4+3sin⁡(4x)64\frac{5 x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}

  3. Add the constant of integration:

    5x16+sin⁡3(2x)48−sin⁡(2x)4+3sin⁡(4x)64+constant\frac{5 x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}+ \mathrm{constant}

The answer is:

5x16+sin⁡3(2x)48−sin⁡(2x)4+3sin⁡(4x)64+constant\frac{5 x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                          
 |                                  3                        
 |    6               sin(2*x)   sin (2*x)   3*sin(4*x)   5*x
 | sin (x)*1 dx = C - -------- + --------- + ---------- + ---
 |                       4           48          64        16
/
3 (sin⁡(4 x)2+2 x)16−sin⁡(2 x)−sin⁡3(2 x)38−3 sin⁡(2 x)8+x42{{{{3\,\left({{\sin \left(4\,x\right)}\over{2}}+2\,x\right)}\over{ 16}}-{{\sin \left(2\,x\right)-{{\sin ^3\left(2\,x\right)}\over{3}} }\over{8}}-{{3\,\sin \left(2\,x\right)}\over{8}}+{{x}\over{4}} }\over{2}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.00.5
Plot the graph f(x)
The answer [src] 
                            3                5          
5    5*cos(1)*sin(1)   5*sin (1)*cos(1)   sin (1)*cos(1)
-- - --------------- - ---------------- - --------------
16          16                24                6
9 sin⁡4+4 sin⁡32−48 sin⁡2+60192{{9\,\sin 4+4\,\sin ^32-48\,\sin 2+60}\over{192}} 
=
= 
                            3                5          
5    5*cos(1)*sin(1)   5*sin (1)*cos(1)   sin (1)*cos(1)
-- - --------------- - ---------------- - --------------
16          16                24                6
−5sin⁡(1)cos⁡(1)16−5sin⁡3(1)cos⁡(1)24−sin⁡5(1)cos⁡(1)6+516- \frac{5 \sin{\left(1 \right)} \cos{\left(1 \right)}}{16} - \frac{5 \sin^{3}{\left(1 \right)} \cos{\left(1 \right)}}{24} - \frac{\sin^{5}{\left(1 \right)} \cos{\left(1 \right)}}{6} + \frac{5}{16}
Simplify
Numerical answer [src] 
0.0653635876732911
0.0653635876732911
The graph
Integral of sin^6x*dx dx

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

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