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tan^3xsec^3x

Integral of tan^3xsec^3x dx

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 |  tan (x)*sec (x) dx
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01tan3(x)sec3(x)dx\int\limits_{0}^{1} \tan^{3}{\left(x \right)} \sec^{3}{\left(x \right)}\, dx
Integral(tan(x)^3*sec(x)^3, (x, 0, 1))
Detail solution
  1. Rewrite the integrand:

    tan3(x)sec3(x)=(sec2(x)1)tan(x)sec3(x)\tan^{3}{\left(x \right)} \sec^{3}{\left(x \right)} = \left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{3}{\left(x \right)}

  2. There are multiple ways to do this integral.

    Method #1

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

      Then let du=tan(x)sec(x)dxdu = \tan{\left(x \right)} \sec{\left(x \right)} dx and substitute dudu:

      (u4u2)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 n1n \neq -1:

          u4du=u55\int u^{4}\, du = \frac{u^{5}}{5}

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

          (u2)du=u2du\int \left(- u^{2}\right)\, du = - \int u^{2}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

            u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

          So, the result is: u33- \frac{u^{3}}{3}

        The result is: u55u33\frac{u^{5}}{5} - \frac{u^{3}}{3}

      Now substitute uu back in:

      sec5(x)5sec3(x)3\frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}

    Method #2

    1. Rewrite the integrand:

      (sec2(x)1)tan(x)sec3(x)=tan(x)sec5(x)tan(x)sec3(x)\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{3}{\left(x \right)} = \tan{\left(x \right)} \sec^{5}{\left(x \right)} - \tan{\left(x \right)} \sec^{3}{\left(x \right)}

    2. Integrate term-by-term:

      1. Let u=sec5(x)u = \sec^{5}{\left(x \right)}.

        Then let du=5tan(x)sec5(x)dxdu = 5 \tan{\left(x \right)} \sec^{5}{\left(x \right)} dx and substitute du5\frac{du}{5}:

        125du\int \frac{1}{25}\, du

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

          15du=1du5\int \frac{1}{5}\, du = \frac{\int 1\, du}{5}

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

            1du=u\int 1\, du = u

          So, the result is: u5\frac{u}{5}

        Now substitute uu back in:

        sec5(x)5\frac{\sec^{5}{\left(x \right)}}{5}

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

        (tan(x)sec3(x))dx=tan(x)sec3(x)dx\int \left(- \tan{\left(x \right)} \sec^{3}{\left(x \right)}\right)\, dx = - \int \tan{\left(x \right)} \sec^{3}{\left(x \right)}\, dx

        1. Let u=sec3(x)u = \sec^{3}{\left(x \right)}.

          Then let du=3tan(x)sec3(x)dxdu = 3 \tan{\left(x \right)} \sec^{3}{\left(x \right)} dx and substitute du3\frac{du}{3}:

          19du\int \frac{1}{9}\, du

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

            13du=1du3\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}

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

              1du=u\int 1\, du = u

            So, the result is: u3\frac{u}{3}

          Now substitute uu back in:

          sec3(x)3\frac{\sec^{3}{\left(x \right)}}{3}

        So, the result is: sec3(x)3- \frac{\sec^{3}{\left(x \right)}}{3}

      The result is: sec5(x)5sec3(x)3\frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}

    Method #3

    1. Rewrite the integrand:

      (sec2(x)1)tan(x)sec3(x)=tan(x)sec5(x)tan(x)sec3(x)\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{3}{\left(x \right)} = \tan{\left(x \right)} \sec^{5}{\left(x \right)} - \tan{\left(x \right)} \sec^{3}{\left(x \right)}

    2. Integrate term-by-term:

      1. Let u=sec5(x)u = \sec^{5}{\left(x \right)}.

        Then let du=5tan(x)sec5(x)dxdu = 5 \tan{\left(x \right)} \sec^{5}{\left(x \right)} dx and substitute du5\frac{du}{5}:

        125du\int \frac{1}{25}\, du

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

          15du=1du5\int \frac{1}{5}\, du = \frac{\int 1\, du}{5}

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

            1du=u\int 1\, du = u

          So, the result is: u5\frac{u}{5}

        Now substitute uu back in:

        sec5(x)5\frac{\sec^{5}{\left(x \right)}}{5}

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

        (tan(x)sec3(x))dx=tan(x)sec3(x)dx\int \left(- \tan{\left(x \right)} \sec^{3}{\left(x \right)}\right)\, dx = - \int \tan{\left(x \right)} \sec^{3}{\left(x \right)}\, dx

        1. Let u=sec3(x)u = \sec^{3}{\left(x \right)}.

          Then let du=3tan(x)sec3(x)dxdu = 3 \tan{\left(x \right)} \sec^{3}{\left(x \right)} dx and substitute du3\frac{du}{3}:

          19du\int \frac{1}{9}\, du

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

            13du=1du3\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}

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

              1du=u\int 1\, du = u

            So, the result is: u3\frac{u}{3}

          Now substitute uu back in:

          sec3(x)3\frac{\sec^{3}{\left(x \right)}}{3}

        So, the result is: sec3(x)3- \frac{\sec^{3}{\left(x \right)}}{3}

      The result is: sec5(x)5sec3(x)3\frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}

  3. Add the constant of integration:

    sec5(x)5sec3(x)3+constant\frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}+ \mathrm{constant}


The answer is:

sec5(x)5sec3(x)3+constant\frac{\sec^{5}{\left(x \right)}}{5} - \frac{\sec^{3}{\left(x \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                          
 |                             3         5   
 |    3       3             sec (x)   sec (x)
 | tan (x)*sec (x) dx = C - ------- + -------
 |                             3         5   
/                                            
5cos2x315cos5x-{{5\,\cos ^2x-3}\over{15\,\cos ^5x}}
The graph
0.001.000.100.200.300.400.500.600.700.800.90-2525
The answer [src]
              2   
2    3 - 5*cos (1)
-- + -------------
15           5    
       15*cos (1) 
13cos31+15cos51+215-{{1}\over{3\,\cos ^31}}+{{1}\over{5\,\cos ^51}}+{{2}\over{15}}
=
=
              2   
2    3 - 5*cos (1)
-- + -------------
15           5    
       15*cos (1) 
215+35cos2(1)15cos5(1)\frac{2}{15} + \frac{3 - 5 \cos^{2}{\left(1 \right)}}{15 \cos^{5}{\left(1 \right)}}
Numerical answer [src]
2.36355929817875
2.36355929817875
The graph
Integral of tan^3xsec^3x dx

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