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sin^5x
Solving integrals/ sin^5x

Integral of sin^5x dx

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

  1. Rewrite the integrand:

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

  2. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

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

        Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

        ∫(−u4) du\int \left(- u^{4}\right)\, du

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

          ∫u4 du=−∫u4 du\int u^{4}\, 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}

          So, the result is: −u55- \frac{u^{5}}{5}

        Now substitute uu back in:

        −cos⁡5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

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

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

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

          Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

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

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

            ∫u2 du=−∫u2 du\int u^{2}\, 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}

            So, the result is: −u33- \frac{u^{3}}{3}

          Now substitute uu back in:

          −cos⁡3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

        So, the result is: 2cos⁡3(x)3\frac{2 \cos^{3}{\left(x \right)}}{3}

      1. The integral of sine is negative cosine:

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

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

    Method #2

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

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

        Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

        ∫(−u4) du\int \left(- u^{4}\right)\, du

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

          ∫u4 du=−∫u4 du\int u^{4}\, 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}

          So, the result is: −u55- \frac{u^{5}}{5}

        Now substitute uu back in:

        −cos⁡5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

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

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

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

          Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

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

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

            ∫u2 du=−∫u2 du\int u^{2}\, 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}

            So, the result is: −u33- \frac{u^{3}}{3}

          Now substitute uu back in:

          −cos⁡3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

        So, the result is: 2cos⁡3(x)3\frac{2 \cos^{3}{\left(x \right)}}{3}

      1. The integral of sine is negative cosine:

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

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

  3. Add the constant of integration:

    −cos⁡5(x)5+2cos⁡3(x)3−cos⁡(x)+constant- \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}+ \mathrm{constant}

The answer is:

−cos⁡5(x)5+2cos⁡3(x)3−cos⁡(x)+constant- \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                             
 |                              5           3   
 |    5                      cos (x)   2*cos (x)
 | sin (x) dx = C - cos(x) - ------- + ---------
 |                              5          3    
/
∫sin⁡5(x) dx=C−cos⁡5(x)5+2cos⁡3(x)3−cos⁡(x)\int \sin^{5}{\left(x \right)}\, dx = C - \frac{\cos^{5}{\left(x \right)}}{5} + \frac{2 \cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.901-1
Plot the graph f(x)
The answer [src] 
                 5           3   
8             cos (1)   2*cos (1)
-- - cos(1) - ------- + ---------
15               5          3
−cos⁡(1)−cos⁡5(1)5+2cos⁡3(1)3+815- \cos{\left(1 \right)} - \frac{\cos^{5}{\left(1 \right)}}{5} + \frac{2 \cos^{3}{\left(1 \right)}}{3} + \frac{8}{15} 
=
= 
                 5           3   
8             cos (1)   2*cos (1)
-- - cos(1) - ------- + ---------
15               5          3
−cos⁡(1)−cos⁡5(1)5+2cos⁡3(1)3+815- \cos{\left(1 \right)} - \frac{\cos^{5}{\left(1 \right)}}{5} + \frac{2 \cos^{3}{\left(1 \right)}}{3} + \frac{8}{15} 
8/15 - cos(1) - cos(1)^5/5 + 2*cos(1)^3/3
Numerical answer [src] 
0.0889743964515759
0.0889743964515759
The graph
Integral of sin^5x dx

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

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