Mister Exam

Other calculators

2^4*sin^8x
Solving integrals/ 2^4*sin^8x

Integral of 2^4*sin^8x dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1              
  /              
 |               
 |   4    8      
 |  2 *sin (x) dx
 |               
/                
0
∫0124sin⁡8(x) dx\int\limits_{0}^{1} 2^{4} \sin^{8}{\left(x \right)}\, dx 
Integral(2^4*sin(x)^8, (x, 0, 1))
Detail solution

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

    ∫24sin⁡8(x) dx=16∫sin⁡8(x) dx\int 2^{4} \sin^{8}{\left(x \right)}\, dx = 16 \int \sin^{8}{\left(x \right)}\, dx

    1. Rewrite the integrand:

      sin⁡8(x)=(12−cos⁡(2x)2)4\sin^{8}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{4}

    2. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

        (12−cos⁡(2x)2)4=cos⁡4(2x)16−cos⁡3(2x)4+3cos⁡2(2x)8−cos⁡(2x)4+116\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{4} = \frac{\cos^{4}{\left(2 x \right)}}{16} - \frac{\cos^{3}{\left(2 x \right)}}{4} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{4} + \frac{1}{16}

      2. Integrate term-by-term:

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

          ∫cos⁡4(2x)16 dx=∫cos⁡4(2x) dx16\int \frac{\cos^{4}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{4}{\left(2 x \right)}\, dx}{16}

          1. Rewrite the integrand:

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

          2. Rewrite the integrand:

            (cos⁡(4x)2+12)2=cos⁡2(4x)4+cos⁡(4x)2+14\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(4 x \right)}}{4} + \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{4}

          3. Integrate term-by-term:

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

              ∫cos⁡2(4x)4 dx=∫cos⁡2(4x) dx4\int \frac{\cos^{2}{\left(4 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(4 x \right)}\, dx}{4}

              1. Rewrite the integrand:

                cos⁡2(4x)=cos⁡(8x)2+12\cos^{2}{\left(4 x \right)} = \frac{\cos{\left(8 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⁡(8x)2 dx=∫cos⁡(8x) dx2\int \frac{\cos{\left(8 x \right)}}{2}\, dx = \frac{\int \cos{\left(8 x \right)}\, dx}{2}

                  1. Let u=8xu = 8 x.

                    Then let du=8dxdu = 8 dx and substitute du8\frac{du}{8}:

                    ∫cos⁡(u)64 du\int \frac{\cos{\left(u \right)}}{64}\, du

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

                      ∫cos⁡(u)8 du=∫cos⁡(u) du8\int \frac{\cos{\left(u \right)}}{8}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}

                      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)8\frac{\sin{\left(u \right)}}{8}

                    Now substitute uu back in:

                    sin⁡(8x)8\frac{\sin{\left(8 x \right)}}{8}

                  So, the result is: sin⁡(8x)16\frac{\sin{\left(8 x \right)}}{16}

                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⁡(8x)16\frac{x}{2} + \frac{\sin{\left(8 x \right)}}{16}

              So, the result is: x8+sin⁡(8x)64\frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64}

            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:

              ∫14 dx=x4\int \frac{1}{4}\, dx = \frac{x}{4}

            The result is: 3x8+sin⁡(4x)8+sin⁡(8x)64\frac{3 x}{8} + \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}

          So, the result is: 3x128+sin⁡(4x)128+sin⁡(8x)1024\frac{3 x}{128} + \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}

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

          ∫(−cos⁡3(2x)4) dx=−∫cos⁡3(2x) dx4\int \left(- \frac{\cos^{3}{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{4}

          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)24−sin⁡(2x)8\frac{\sin^{3}{\left(2 x \right)}}{24} - \frac{\sin{\left(2 x \right)}}{8}

        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:

          ∫(−cos⁡(2x)4) dx=−∫cos⁡(2x) dx4\int \left(- \frac{\cos{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{4}

          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: −sin⁡(2x)8- \frac{\sin{\left(2 x \right)}}{8}

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

          ∫116 dx=x16\int \frac{1}{16}\, dx = \frac{x}{16}

        The result is: 35x128+sin⁡3(2x)24−sin⁡(2x)4+7sin⁡(4x)128+sin⁡(8x)1024\frac{35 x}{128} + \frac{\sin^{3}{\left(2 x \right)}}{24} - \frac{\sin{\left(2 x \right)}}{4} + \frac{7 \sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}

      Method #2

      1. Rewrite the integrand:

        (12−cos⁡(2x)2)4=cos⁡4(2x)16−cos⁡3(2x)4+3cos⁡2(2x)8−cos⁡(2x)4+116\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{4} = \frac{\cos^{4}{\left(2 x \right)}}{16} - \frac{\cos^{3}{\left(2 x \right)}}{4} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{4} + \frac{1}{16}

      2. Integrate term-by-term:

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

          ∫cos⁡4(2x)16 dx=∫cos⁡4(2x) dx16\int \frac{\cos^{4}{\left(2 x \right)}}{16}\, dx = \frac{\int \cos^{4}{\left(2 x \right)}\, dx}{16}

          1. Rewrite the integrand:

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

          2. Rewrite the integrand:

            (cos⁡(4x)2+12)2=cos⁡2(4x)4+cos⁡(4x)2+14\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}\right)^{2} = \frac{\cos^{2}{\left(4 x \right)}}{4} + \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{4}

          3. Integrate term-by-term:

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

              ∫cos⁡2(4x)4 dx=∫cos⁡2(4x) dx4\int \frac{\cos^{2}{\left(4 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(4 x \right)}\, dx}{4}

              1. Rewrite the integrand:

                cos⁡2(4x)=cos⁡(8x)2+12\cos^{2}{\left(4 x \right)} = \frac{\cos{\left(8 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⁡(8x)2 dx=∫cos⁡(8x) dx2\int \frac{\cos{\left(8 x \right)}}{2}\, dx = \frac{\int \cos{\left(8 x \right)}\, dx}{2}

                  1. Let u=8xu = 8 x.

                    Then let du=8dxdu = 8 dx and substitute du8\frac{du}{8}:

                    ∫cos⁡(u)64 du\int \frac{\cos{\left(u \right)}}{64}\, du

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

                      ∫cos⁡(u)8 du=∫cos⁡(u) du8\int \frac{\cos{\left(u \right)}}{8}\, du = \frac{\int \cos{\left(u \right)}\, du}{8}

                      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)8\frac{\sin{\left(u \right)}}{8}

                    Now substitute uu back in:

                    sin⁡(8x)8\frac{\sin{\left(8 x \right)}}{8}

                  So, the result is: sin⁡(8x)16\frac{\sin{\left(8 x \right)}}{16}

                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⁡(8x)16\frac{x}{2} + \frac{\sin{\left(8 x \right)}}{16}

              So, the result is: x8+sin⁡(8x)64\frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64}

            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:

              ∫14 dx=x4\int \frac{1}{4}\, dx = \frac{x}{4}

            The result is: 3x8+sin⁡(4x)8+sin⁡(8x)64\frac{3 x}{8} + \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}

          So, the result is: 3x128+sin⁡(4x)128+sin⁡(8x)1024\frac{3 x}{128} + \frac{\sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}

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

          ∫(−cos⁡3(2x)4) dx=−∫cos⁡3(2x) dx4\int \left(- \frac{\cos^{3}{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{4}

          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)24−sin⁡(2x)8\frac{\sin^{3}{\left(2 x \right)}}{24} - \frac{\sin{\left(2 x \right)}}{8}

        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:

          ∫(−cos⁡(2x)4) dx=−∫cos⁡(2x) dx4\int \left(- \frac{\cos{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{4}

          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: −sin⁡(2x)8- \frac{\sin{\left(2 x \right)}}{8}

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

          ∫116 dx=x16\int \frac{1}{16}\, dx = \frac{x}{16}

        The result is: 35x128+sin⁡3(2x)24−sin⁡(2x)4+7sin⁡(4x)128+sin⁡(8x)1024\frac{35 x}{128} + \frac{\sin^{3}{\left(2 x \right)}}{24} - \frac{\sin{\left(2 x \right)}}{4} + \frac{7 \sin{\left(4 x \right)}}{128} + \frac{\sin{\left(8 x \right)}}{1024}

    So, the result is: 35x8+2sin⁡3(2x)3−4sin⁡(2x)+7sin⁡(4x)8+sin⁡(8x)64\frac{35 x}{8} + \frac{2 \sin^{3}{\left(2 x \right)}}{3} - 4 \sin{\left(2 x \right)} + \frac{7 \sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}

  2. Add the constant of integration:

    35x8+2sin⁡3(2x)3−4sin⁡(2x)+7sin⁡(4x)8+sin⁡(8x)64+constant\frac{35 x}{8} + \frac{2 \sin^{3}{\left(2 x \right)}}{3} - 4 \sin{\left(2 x \right)} + \frac{7 \sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}+ \mathrm{constant}

The answer is:

35x8+2sin⁡3(2x)3−4sin⁡(2x)+7sin⁡(4x)8+sin⁡(8x)64+constant\frac{35 x}{8} + \frac{2 \sin^{3}{\left(2 x \right)}}{3} - 4 \sin{\left(2 x \right)} + \frac{7 \sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                           
 |                                                  3                         
 |  4    8                          sin(8*x)   2*sin (2*x)   7*sin(4*x)   35*x
 | 2 *sin (x) dx = C - 4*sin(2*x) + -------- + ----------- + ---------- + ----
 |                                     64           3            8         8  
/
8 (sin⁡(8 x)2+4 x8+sin⁡(4 x)2+x32+3 (sin⁡(4 x)2+2 x)16−sin⁡(2 x)−sin⁡3(2 x)34−sin⁡(2 x)4+x8)8\,\left({{{{{{\sin \left(8\,x\right)}\over{2}}+4\,x}\over{8}}+{{ \sin \left(4\,x\right)}\over{2}}+x}\over{32}}+{{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{4}}-{{\sin \left( 2\,x\right)}\over{4}}+{{x}\over{8}}\right)
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9005
Plot the graph f(x)
The answer [src] 
                                                 3                  5          
35        7             35*cos(1)*sin(1)   35*sin (1)*cos(1)   7*sin (1)*cos(1)
-- - 2*sin (1)*cos(1) - ---------------- - ----------------- - ----------------
8                              8                   12                 3
3 sin⁡8+168 sin⁡4+128 sin⁡32−768 sin⁡2+840192{{3\,\sin 8+168\,\sin 4+128\,\sin ^32-768\,\sin 2+840}\over{192}} 
=
= 
                                                 3                  5          
35        7             35*cos(1)*sin(1)   35*sin (1)*cos(1)   7*sin (1)*cos(1)
-- - 2*sin (1)*cos(1) - ---------------- - ----------------- - ----------------
8                              8                   12                 3
−35sin⁡(1)cos⁡(1)8−35sin⁡3(1)cos⁡(1)12−7sin⁡5(1)cos⁡(1)3−2sin⁡7(1)cos⁡(1)+358- \frac{35 \sin{\left(1 \right)} \cos{\left(1 \right)}}{8} - \frac{35 \sin^{3}{\left(1 \right)} \cos{\left(1 \right)}}{12} - \frac{7 \sin^{5}{\left(1 \right)} \cos{\left(1 \right)}}{3} - 2 \sin^{7}{\left(1 \right)} \cos{\left(1 \right)} + \frac{35}{8}
Simplify
Numerical answer [src] 
0.592284795018988
0.592284795018988
The graph
Integral of 2^4*sin^8x dx

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

    © Mister Exam – Calculator