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tan^3(x)sec^7(x)

Integral of tan^3(x)sec^7(x) dx

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 |     3       7      
 |  tan (x)*sec (x) dx
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01tan3(x)sec7(x)dx\int\limits_{0}^{1} \tan^{3}{\left(x \right)} \sec^{7}{\left(x \right)}\, dx
Integral(tan(x)^3*sec(x)^7, (x, 0, 1))
Detail solution
  1. Rewrite the integrand:

    tan3(x)sec7(x)=(sec2(x)1)tan(x)sec7(x)\tan^{3}{\left(x \right)} \sec^{7}{\left(x \right)} = \left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{7}{\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:

      (u8u6)du\int \left(u^{8} - u^{6}\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:

          u8du=u99\int u^{8}\, du = \frac{u^{9}}{9}

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

          (u6)du=u6du\int \left(- u^{6}\right)\, du = - \int u^{6}\, du

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

            u6du=u77\int u^{6}\, du = \frac{u^{7}}{7}

          So, the result is: u77- \frac{u^{7}}{7}

        The result is: u99u77\frac{u^{9}}{9} - \frac{u^{7}}{7}

      Now substitute uu back in:

      sec9(x)9sec7(x)7\frac{\sec^{9}{\left(x \right)}}{9} - \frac{\sec^{7}{\left(x \right)}}{7}

    Method #2

    1. Rewrite the integrand:

      (sec2(x)1)tan(x)sec7(x)=tan(x)sec9(x)tan(x)sec7(x)\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{7}{\left(x \right)} = \tan{\left(x \right)} \sec^{9}{\left(x \right)} - \tan{\left(x \right)} \sec^{7}{\left(x \right)}

    2. Integrate term-by-term:

      1. Let u=sec9(x)u = \sec^{9}{\left(x \right)}.

        Then let du=9tan(x)sec9(x)dxdu = 9 \tan{\left(x \right)} \sec^{9}{\left(x \right)} dx and substitute du9\frac{du}{9}:

        181du\int \frac{1}{81}\, du

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

          19du=1du9\int \frac{1}{9}\, du = \frac{\int 1\, du}{9}

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

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

          So, the result is: u9\frac{u}{9}

        Now substitute uu back in:

        sec9(x)9\frac{\sec^{9}{\left(x \right)}}{9}

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

        (tan(x)sec7(x))dx=tan(x)sec7(x)dx\int \left(- \tan{\left(x \right)} \sec^{7}{\left(x \right)}\right)\, dx = - \int \tan{\left(x \right)} \sec^{7}{\left(x \right)}\, dx

        1. Let u=sec7(x)u = \sec^{7}{\left(x \right)}.

          Then let du=7tan(x)sec7(x)dxdu = 7 \tan{\left(x \right)} \sec^{7}{\left(x \right)} dx and substitute du7\frac{du}{7}:

          149du\int \frac{1}{49}\, du

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

            17du=1du7\int \frac{1}{7}\, du = \frac{\int 1\, du}{7}

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

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

            So, the result is: u7\frac{u}{7}

          Now substitute uu back in:

          sec7(x)7\frac{\sec^{7}{\left(x \right)}}{7}

        So, the result is: sec7(x)7- \frac{\sec^{7}{\left(x \right)}}{7}

      The result is: sec9(x)9sec7(x)7\frac{\sec^{9}{\left(x \right)}}{9} - \frac{\sec^{7}{\left(x \right)}}{7}

    Method #3

    1. Rewrite the integrand:

      (sec2(x)1)tan(x)sec7(x)=tan(x)sec9(x)tan(x)sec7(x)\left(\sec^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} \sec^{7}{\left(x \right)} = \tan{\left(x \right)} \sec^{9}{\left(x \right)} - \tan{\left(x \right)} \sec^{7}{\left(x \right)}

    2. Integrate term-by-term:

      1. Let u=sec9(x)u = \sec^{9}{\left(x \right)}.

        Then let du=9tan(x)sec9(x)dxdu = 9 \tan{\left(x \right)} \sec^{9}{\left(x \right)} dx and substitute du9\frac{du}{9}:

        181du\int \frac{1}{81}\, du

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

          19du=1du9\int \frac{1}{9}\, du = \frac{\int 1\, du}{9}

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

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

          So, the result is: u9\frac{u}{9}

        Now substitute uu back in:

        sec9(x)9\frac{\sec^{9}{\left(x \right)}}{9}

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

        (tan(x)sec7(x))dx=tan(x)sec7(x)dx\int \left(- \tan{\left(x \right)} \sec^{7}{\left(x \right)}\right)\, dx = - \int \tan{\left(x \right)} \sec^{7}{\left(x \right)}\, dx

        1. Let u=sec7(x)u = \sec^{7}{\left(x \right)}.

          Then let du=7tan(x)sec7(x)dxdu = 7 \tan{\left(x \right)} \sec^{7}{\left(x \right)} dx and substitute du7\frac{du}{7}:

          149du\int \frac{1}{49}\, du

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

            17du=1du7\int \frac{1}{7}\, du = \frac{\int 1\, du}{7}

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

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

            So, the result is: u7\frac{u}{7}

          Now substitute uu back in:

          sec7(x)7\frac{\sec^{7}{\left(x \right)}}{7}

        So, the result is: sec7(x)7- \frac{\sec^{7}{\left(x \right)}}{7}

      The result is: sec9(x)9sec7(x)7\frac{\sec^{9}{\left(x \right)}}{9} - \frac{\sec^{7}{\left(x \right)}}{7}

  3. Add the constant of integration:

    sec9(x)9sec7(x)7+constant\frac{\sec^{9}{\left(x \right)}}{9} - \frac{\sec^{7}{\left(x \right)}}{7}+ \mathrm{constant}


The answer is:

sec9(x)9sec7(x)7+constant\frac{\sec^{9}{\left(x \right)}}{9} - \frac{\sec^{7}{\left(x \right)}}{7}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                          
 |                             7         9   
 |    3       7             sec (x)   sec (x)
 | tan (x)*sec (x) dx = C - ------- + -------
 |                             7         9   
/                                            
9cos2x763cos9x-{{9\,\cos ^2x-7}\over{63\,\cos ^9x}}
The graph
0.001.000.100.200.300.400.500.600.700.800.90-500500
The answer [src]
              2   
2    7 - 9*cos (1)
-- + -------------
63           9    
       63*cos (1) 
17cos71+19cos91+263-{{1}\over{7\,\cos ^71}}+{{1}\over{9\,\cos ^91}}+{{2}\over{63}}
=
=
              2   
2    7 - 9*cos (1)
-- + -------------
63           9    
       63*cos (1) 
263+79cos2(1)63cos9(1)\frac{2}{63} + \frac{7 - 9 \cos^{2}{\left(1 \right)}}{63 \cos^{9}{\left(1 \right)}}
Numerical answer [src]
17.7195471832138
17.7195471832138
The graph
Integral of tan^3(x)sec^7(x) dx

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