Mister Exam

Other calculators

tan^5x+tan^3x
Solving integrals/ tan^5x+tan^3x

Integral of tan^5x+tan^3x dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1                       
  /                       
 |                        
 |  /   5         3   \   
 |  \tan (x) + tan (x)/ dx
 |                        
/                         
0
∫01(tan⁡5(x)+tan⁡3(x)) dx\int\limits_{0}^{1} \left(\tan^{5}{\left(x \right)} + \tan^{3}{\left(x \right)}\right)\, dx 
Integral(tan(x)^5 + tan(x)^3, (x, 0, 1))
Detail solution

  1. Integrate term-by-term:

    1. Rewrite the integrand:

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

    1. Rewrite the integrand:

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

    2. 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}:

      ∫u−14u du\int \frac{u - 1}{4 u}\, du

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

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

        1. Rewrite the integrand:

          u−1u=1−1u\frac{u - 1}{u} = 1 - \frac{1}{u}

        2. Integrate term-by-term:

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

            ∫1 du=u\int 1\, du = u

          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)}

          The result is: u−log⁡(u)u - \log{\left(u \right)}

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

      Now substitute uu back in:

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

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

  2. Now simplify:

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

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                              
 |                                 2         4   
 | /   5         3   \          sec (x)   sec (x)
 | \tan (x) + tan (x)/ dx = C - ------- + -------
 |                                 2         4   
/
4 sin⁡2x−34 sin⁡4x−8 sin⁡2x+4−12 sin⁡2x−2{{4\,\sin ^2x-3}\over{4\,\sin ^4x-8\,\sin ^2x+4}}-{{1}\over{2\, \sin ^2x-2}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9020-10
Plot the graph f(x)
The answer [src] 
                          2   
1       1       -1 + 4*cos (1)
- + --------- - --------------
4        2             4      
    2*cos (1)     4*cos (1)
sin⁡414 sin⁡41−8 sin⁡21+4{{\sin ^41}\over{4\,\sin ^41-8\,\sin ^21+4}} 
=
= 
                          2   
1       1       -1 + 4*cos (1)
- + --------- - --------------
4        2             4      
    2*cos (1)     4*cos (1)
−−1+4cos⁡2(1)4cos⁡4(1)+14+12cos⁡2(1)- \frac{-1 + 4 \cos^{2}{\left(1 \right)}}{4 \cos^{4}{\left(1 \right)}} + \frac{1}{4} + \frac{1}{2 \cos^{2}{\left(1 \right)}}
Simplify
Numerical answer [src] 
1.47078538753166
1.47078538753166
The graph
Integral of tan^5x+tan^3x dx

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

    © Mister Exam – Calculator