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sin^5x/cosx
Solving integrals/ sin^5x/cosx

Integral of sin^5x/cosx dx

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

  1. Rewrite the integrand:

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

  2. There are multiple ways to do this integral.

    Method #1

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

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

      ∫(−u2−2u+12u) du\int \left(- \frac{u^{2} - 2 u + 1}{2 u}\right)\, du

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

        ∫u2−2u+1u du=−∫u2−2u+1u du2\int \frac{u^{2} - 2 u + 1}{u}\, du = - \frac{\int \frac{u^{2} - 2 u + 1}{u}\, du}{2}

        1. Rewrite the integrand:

          u2−2u+1u=u−2+1u\frac{u^{2} - 2 u + 1}{u} = u - 2 + \frac{1}{u}

        2. 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:

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

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

            ∫(−2) du=−2u\int \left(-2\right)\, du = - 2 u

          1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

          The result is: u22−2u+log⁡(u)\frac{u^{2}}{2} - 2 u + \log{\left(u \right)}

        So, the result is: −u24+u−log⁡(u)2- \frac{u^{2}}{4} + u - \frac{\log{\left(u \right)}}{2}

      Now substitute uu back in:

      −log⁡(cos⁡2(x))2−cos⁡4(x)4+cos⁡2(x)- \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{2} - \frac{\cos^{4}{\left(x \right)}}{4} + \cos^{2}{\left(x \right)}

    Method #2

    1. Rewrite the integrand:

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

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

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

      ∫(−u4−2u2+1u) du\int \left(- \frac{u^{4} - 2 u^{2} + 1}{u}\right)\, du

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

        ∫u4−2u2+1u du=−∫u4−2u2+1u du\int \frac{u^{4} - 2 u^{2} + 1}{u}\, du = - \int \frac{u^{4} - 2 u^{2} + 1}{u}\, du

        1. Let u=u2u = u^{2}.

          Then let du=2ududu = 2 u du and substitute du2\frac{du}{2}:

          ∫u2−2u+12u du\int \frac{u^{2} - 2 u + 1}{2 u}\, du

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

            ∫u2−2u+1u du=∫u2−2u+1u du2\int \frac{u^{2} - 2 u + 1}{u}\, du = \frac{\int \frac{u^{2} - 2 u + 1}{u}\, du}{2}

            1. Rewrite the integrand:

              u2−2u+1u=u−2+1u\frac{u^{2} - 2 u + 1}{u} = u - 2 + \frac{1}{u}

            2. 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:

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

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

                ∫(−2) du=−2u\int \left(-2\right)\, du = - 2 u

              1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

              The result is: u22−2u+log⁡(u)\frac{u^{2}}{2} - 2 u + \log{\left(u \right)}

            So, the result is: u24−u+log⁡(u)2\frac{u^{2}}{4} - u + \frac{\log{\left(u \right)}}{2}

          Now substitute uu back in:

          u44−u2+log⁡(u2)2\frac{u^{4}}{4} - u^{2} + \frac{\log{\left(u^{2} \right)}}{2}

        So, the result is: −u44+u2−log⁡(u2)2- \frac{u^{4}}{4} + u^{2} - \frac{\log{\left(u^{2} \right)}}{2}

      Now substitute uu back in:

      −log⁡(cos⁡2(x))2−cos⁡4(x)4+cos⁡2(x)- \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{2} - \frac{\cos^{4}{\left(x \right)}}{4} + \cos^{2}{\left(x \right)}

    Method #3

    1. Rewrite the integrand:

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

        ∫(−u3) du\int \left(- u^{3}\right)\, du

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

          ∫u3 du=−∫u3 du\int u^{3}\, du = - \int u^{3}\, du

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

            ∫u3 du=u44\int u^{3}\, du = \frac{u^{4}}{4}

          So, the result is: −u44- \frac{u^{4}}{4}

        Now substitute uu back in:

        −cos⁡4(x)4- \frac{\cos^{4}{\left(x \right)}}{4}

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

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

          ∫(−u) du\int \left(- u\right)\, du

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

            ∫u du=−∫u du\int u\, du = - \int u\, du

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

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

            So, the result is: −u22- \frac{u^{2}}{2}

          Now substitute uu back in:

          −cos⁡2(x)2- \frac{\cos^{2}{\left(x \right)}}{2}

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

      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:

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

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

          ∫1u du=−∫1u du\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du

          1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

          So, the result is: −log⁡(u)- \log{\left(u \right)}

        Now substitute uu back in:

        −log⁡(cos⁡(x))- \log{\left(\cos{\left(x \right)} \right)}

      The result is: −log⁡(cos⁡(x))−cos⁡4(x)4+cos⁡2(x)- \log{\left(\cos{\left(x \right)} \right)} - \frac{\cos^{4}{\left(x \right)}}{4} + \cos^{2}{\left(x \right)}

  3. Add the constant of integration:

    −log⁡(cos⁡2(x))2−cos⁡4(x)4+cos⁡2(x)+constant- \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{2} - \frac{\cos^{4}{\left(x \right)}}{4} + \cos^{2}{\left(x \right)}+ \mathrm{constant}

The answer is:

−log⁡(cos⁡2(x))2−cos⁡4(x)4+cos⁡2(x)+constant- \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{2} - \frac{\cos^{4}{\left(x \right)}}{4} + \cos^{2}{\left(x \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                 
 |                                                  
 |    5                          /   2   \      4   
 | sin (x)             2      log\cos (x)/   cos (x)
 | ------- dx = C + cos (x) - ------------ - -------
 |  cos(x)                         2            4   
 |                                                  
/
∫sin⁡5(x)cos⁡(x) dx=C−log⁡(cos⁡2(x))2−cos⁡4(x)4+cos⁡2(x)\int \frac{\sin^{5}{\left(x \right)}}{\cos{\left(x \right)}}\, dx = C - \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{2} - \frac{\cos^{4}{\left(x \right)}}{4} + \cos^{2}{\left(x \right)}
Simplify
The graph
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Plot the graph f(x)
The answer [src] 
                                 4   
  3      2                    cos (1)
- - + cos (1) - log(cos(1)) - -------
  4                              4
−34−cos⁡4(1)4+cos⁡2(1)−log⁡(cos⁡(1))- \frac{3}{4} - \frac{\cos^{4}{\left(1 \right)}}{4} + \cos^{2}{\left(1 \right)} - \log{\left(\cos{\left(1 \right)} \right)} 
=
= 
                                 4   
  3      2                    cos (1)
- - + cos (1) - log(cos(1)) - -------
  4                              4
−34−cos⁡4(1)4+cos⁡2(1)−log⁡(cos⁡(1))- \frac{3}{4} - \frac{\cos^{4}{\left(1 \right)}}{4} + \cos^{2}{\left(1 \right)} - \log{\left(\cos{\left(1 \right)} \right)} 
-3/4 + cos(1)^2 - log(cos(1)) - cos(1)^4/4
Numerical answer [src] 
0.136247769832824
0.136247769832824
The graph
Integral of sin^5x/cosx dx

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

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