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tan^5xdx
Solving integrals/ tan^5xdx

Integral of tan^5xdx dx

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

  1. Rewrite the integrand:

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

  2. There are multiple ways to do this integral.

    Method #1

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

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

      ∫u2−2u+14u du\int \frac{u^{2} - 2 u + 1}{4 u}\, du

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

        ∫u2−2u+12u du=∫u2−2u+1u du2\int \frac{u^{2} - 2 u + 1}{2 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⁡(sec⁡2(x))2+sec⁡4(x)4−sec⁡2(x)\frac{\log{\left(\sec^{2}{\left(x \right)} \right)}}{2} + \frac{\sec^{4}{\left(x \right)}}{4} - \sec^{2}{\left(x \right)}

    Method #2

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

      1. Let u=sec⁡4(x)u = \sec^{4}{\left(x \right)}.

        Then let du=4tan⁡(x)sec⁡4(x)dxdu = 4 \tan{\left(x \right)} \sec^{4}{\left(x \right)} dx and substitute du4\frac{du}{4}:

        ∫116 du\int \frac{1}{16}\, du

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

          ∫14 du=∫1 du4\int \frac{1}{4}\, du = \frac{\int 1\, du}{4}

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

            ∫1 du=u\int 1\, du = u

          So, the result is: u4\frac{u}{4}

        Now substitute uu back in:

        sec⁡4(x)4\frac{\sec^{4}{\left(x \right)}}{4}

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

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

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

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

          ∫14 du\int \frac{1}{4}\, du

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

            ∫12 du=∫1 du2\int \frac{1}{2}\, du = \frac{\int 1\, du}{2}

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

              ∫1 du=u\int 1\, du = u

            So, the result is: u2\frac{u}{2}

          Now substitute uu back in:

          sec⁡2(x)2\frac{\sec^{2}{\left(x \right)}}{2}

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

      1. Rewrite the integrand:

        tan⁡(x)=sin⁡(x)cos⁡(x)\tan{\left(x \right)} = \frac{\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:

        ∫1u du\int \frac{1}{u}\, du

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

          ∫(−1u) du=−∫1u du\int \left(- \frac{1}{u}\right)\, 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))+sec⁡4(x)4−sec⁡2(x)- \log{\left(\cos{\left(x \right)} \right)} + \frac{\sec^{4}{\left(x \right)}}{4} - \sec^{2}{\left(x \right)}

    Method #3

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

      1. Let u=sec⁡4(x)u = \sec^{4}{\left(x \right)}.

        Then let du=4tan⁡(x)sec⁡4(x)dxdu = 4 \tan{\left(x \right)} \sec^{4}{\left(x \right)} dx and substitute du4\frac{du}{4}:

        ∫116 du\int \frac{1}{16}\, du

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

          ∫14 du=∫1 du4\int \frac{1}{4}\, du = \frac{\int 1\, du}{4}

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

            ∫1 du=u\int 1\, du = u

          So, the result is: u4\frac{u}{4}

        Now substitute uu back in:

        sec⁡4(x)4\frac{\sec^{4}{\left(x \right)}}{4}

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

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

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

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

          ∫14 du\int \frac{1}{4}\, du

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

            ∫12 du=∫1 du2\int \frac{1}{2}\, du = \frac{\int 1\, du}{2}

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

              ∫1 du=u\int 1\, du = u

            So, the result is: u2\frac{u}{2}

          Now substitute uu back in:

          sec⁡2(x)2\frac{\sec^{2}{\left(x \right)}}{2}

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

      1. Rewrite the integrand:

        tan⁡(x)=sin⁡(x)cos⁡(x)\tan{\left(x \right)} = \frac{\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:

        ∫1u du\int \frac{1}{u}\, du

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

          ∫(−1u) du=−∫1u du\int \left(- \frac{1}{u}\right)\, 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))+sec⁡4(x)4−sec⁡2(x)- \log{\left(\cos{\left(x \right)} \right)} + \frac{\sec^{4}{\left(x \right)}}{4} - \sec^{2}{\left(x \right)}

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                                   
 |                       /   2   \                4   
 |    5               log\sec (x)/      2      sec (x)
 | tan (x)*1 dx = C + ------------ - sec (x) + -------
 |                         2                      4   
/
4 sin⁡2x−34 sin⁡4x−8 sin⁡2x+4−log⁡(sin⁡2x−1)2{{4\,\sin ^2x-3}\over{4\,\sin ^4x-8\,\sin ^2x+4}}-{{\log \left( \sin ^2x-1\right)}\over{2}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-1010
Plot the graph f(x)
The answer [src] 
                            2   
3                 -1 + 4*cos (1)
- - log(cos(1)) - --------------
4                        4      
                    4*cos (1)
−log⁡(1−sin⁡21)2−34 sin⁡41−8 sin⁡21+4+sin⁡21sin⁡41−2 sin⁡21+1+34-{{\log \left(1-\sin ^21\right)}\over{2}}-{{3}\over{4\,\sin ^41-8\, \sin ^21+4}}+{{\sin ^21}\over{\sin ^41-2\,\sin ^21+1}}+{{3}\over{4}} 
=
= 
                            2   
3                 -1 + 4*cos (1)
- - log(cos(1)) - --------------
4                        4      
                    4*cos (1)
−−1+4cos⁡2(1)4cos⁡4(1)−log⁡(cos⁡(1))+34- \frac{-1 + 4 \cos^{2}{\left(1 \right)}}{4 \cos^{4}{\left(1 \right)}} - \log{\left(\cos{\left(1 \right)} \right)} + \frac{3}{4}
Simplify
Numerical answer [src] 
0.87365244751029
0.87365244751029
The graph
Integral of tan^5xdx dx

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

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