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dx/cos^4x
Solving integrals/ dx/cos^4x

Integral of dx/cos^4x dx

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∫011⋅1cos⁡4(x) dx\int\limits_{0}^{1} 1 \cdot \frac{1}{\cos^{4}{\left(x \right)}}\, dx 
Integral(1/cos(x)^4, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

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

      ∫(u2+1) du\int \left(u^{2} + 1\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:

          ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

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

          ∫1 du=u\int 1\, du = u

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

      Now substitute uu back in:

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

    Method #2

    1. Rewrite the integrand:

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

        ∫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}

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

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

    Method #3

    1. Rewrite the integrand:

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

        ∫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}

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

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

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                   
 |                       3            
 |      1             tan (x)         
 | 1*------- dx = C + ------- + tan(x)
 |      4                3            
 |   cos (x)                          
 |                                    
/
tan⁡3x3+tan⁡x{{\tan ^3x}\over{3}}+\tan x
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90020
Plot the graph f(x)
The answer [src] 
  sin(1)    2*sin(1)
--------- + --------
     3      3*cos(1)
3*cos (1)
tan⁡313+tan⁡1{{\tan ^31}\over{3}}+\tan 1 
=
= 
  sin(1)    2*sin(1)
--------- + --------
     3      3*cos(1)
3*cos (1)
2sin⁡(1)3cos⁡(1)+sin⁡(1)3cos⁡3(1)\frac{2 \sin{\left(1 \right)}}{3 \cos{\left(1 \right)}} + \frac{\sin{\left(1 \right)}}{3 \cos^{3}{\left(1 \right)}}
Упростить
Numerical answer [src] 
2.81658164059915
2.81658164059915
The graph
Integral of dx/cos^4x dx

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

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