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tan^2(x)sec^4(x)
Solving integrals/ tan^2(x)sec^4(x)

Integral of tan^2(x)sec^4(x) dx

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∫01tan⁡2(x)sec⁡4(x) dx\int\limits_{0}^{1} \tan^{2}{\left(x \right)} \sec^{4}{\left(x \right)}\, dx 
Integral(tan(x)^2*sec(x)^4, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

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

  2. There are multiple ways to do this integral.

    Method #1

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

      Then let du=(tan⁡2(x)+1)dxdu = \left(\tan^{2}{\left(x \right)} + 1\right) dx and substitute dudu:

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

      1. 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:

          ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

        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}

        The result is: u55+u33\frac{u^{5}}{5} + \frac{u^{3}}{3}

      Now substitute uu back in:

      tan⁡5(x)5+tan⁡3(x)3\frac{\tan^{5}{\left(x \right)}}{5} + \frac{\tan^{3}{\left(x \right)}}{3}

    Method #2

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

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

        Then let du=(tan⁡2(x)+1)dxdu = \left(\tan^{2}{\left(x \right)} + 1\right) dx and substitute dudu:

        ∫u4 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}

        Now substitute uu back in:

        tan⁡5(x)5\frac{\tan^{5}{\left(x \right)}}{5}

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

        Then let du=(tan⁡2(x)+1)dxdu = \left(\tan^{2}{\left(x \right)} + 1\right) dx and substitute dudu:

        ∫u2 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}

        Now substitute uu back in:

        tan⁡3(x)3\frac{\tan^{3}{\left(x \right)}}{3}

      The result is: tan⁡5(x)5+tan⁡3(x)3\frac{\tan^{5}{\left(x \right)}}{5} + \frac{\tan^{3}{\left(x \right)}}{3}

    Method #3

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

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

        Then let du=(tan⁡2(x)+1)dxdu = \left(\tan^{2}{\left(x \right)} + 1\right) dx and substitute dudu:

        ∫u4 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}

        Now substitute uu back in:

        tan⁡5(x)5\frac{\tan^{5}{\left(x \right)}}{5}

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

        Then let du=(tan⁡2(x)+1)dxdu = \left(\tan^{2}{\left(x \right)} + 1\right) dx and substitute dudu:

        ∫u2 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}

        Now substitute uu back in:

        tan⁡3(x)3\frac{\tan^{3}{\left(x \right)}}{3}

      The result is: tan⁡5(x)5+tan⁡3(x)3\frac{\tan^{5}{\left(x \right)}}{5} + \frac{\tan^{3}{\left(x \right)}}{3}

  3. Add the constant of integration:

    tan⁡5(x)5+tan⁡3(x)3+constant\frac{\tan^{5}{\left(x \right)}}{5} + \frac{\tan^{3}{\left(x \right)}}{3}+ \mathrm{constant}

The answer is:

tan⁡5(x)5+tan⁡3(x)3+constant\frac{\tan^{5}{\left(x \right)}}{5} + \frac{\tan^{3}{\left(x \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                          
 |                             3         5   
 |    2       4             tan (x)   tan (x)
 | tan (x)*sec (x) dx = C + ------- + -------
 |                             3         5   
/
3 tan⁡5x+5 tan⁡3x15{{3\,\tan ^5x+5\,\tan ^3x}\over{15}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90050
Plot the graph f(x)
The answer [src] 
   2*sin(1)     sin(1)       sin(1) 
- --------- - ---------- + ---------
  15*cos(1)         3           5   
              15*cos (1)   5*cos (1)
3 tan⁡51+5 tan⁡3115{{3\,\tan ^51+5\,\tan ^31}\over{15}} 
=
= 
   2*sin(1)     sin(1)       sin(1) 
- --------- - ---------- + ---------
  15*cos(1)         3           5   
              15*cos (1)   5*cos (1)
−sin⁡(1)15cos⁡3(1)−2sin⁡(1)15cos⁡(1)+sin⁡(1)5cos⁡5(1)- \frac{\sin{\left(1 \right)}}{15 \cos^{3}{\left(1 \right)}} - \frac{2 \sin{\left(1 \right)}}{15 \cos{\left(1 \right)}} + \frac{\sin{\left(1 \right)}}{5 \cos^{5}{\left(1 \right)}}
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Numerical answer [src] 
3.09166393502534
3.09166393502534
The graph
Integral of tan^2(x)sec^4(x) dx

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

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