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sin^3x/cos^4x
Solving integrals/ sin^3x/cos^4x

Integral of sin^3x/cos^4x dx

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0
∫01sin⁡3(x)cos⁡4(x) dx\int\limits_{0}^{1} \frac{\sin^{3}{\left(x \right)}}{\cos^{4}{\left(x \right)}}\, dx
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Detail solution

  1. Rewrite the integrand:

    sin⁡3(x)cos⁡4(x)=(1−cos⁡2(x))sin⁡(x)cos⁡4(x)\frac{\sin^{3}{\left(x \right)}}{\cos^{4}{\left(x \right)}} = \frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos^{4}{\left(x \right)}}

  2. There are multiple ways to do this integral.

    Method #1

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

      Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute dudu:

      ∫u2−1u4 du\int \frac{u^{2} - 1}{u^{4}}\, du

      1. Rewrite the integrand:

        u2−1u4=1u2−1u4\frac{u^{2} - 1}{u^{4}} = \frac{1}{u^{2}} - \frac{1}{u^{4}}

      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:

          ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

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

          ∫(−1u4) du=−∫1u4 du\int \left(- \frac{1}{u^{4}}\right)\, du = - \int \frac{1}{u^{4}}\, du

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

            ∫1u4 du=−13u3\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}

          So, the result is: 13u3\frac{1}{3 u^{3}}

        The result is: −1u+13u3- \frac{1}{u} + \frac{1}{3 u^{3}}

      Now substitute uu back in:

      −1cos⁡(x)+13cos⁡3(x)- \frac{1}{\cos{\left(x \right)}} + \frac{1}{3 \cos^{3}{\left(x \right)}}

    Method #2

    1. Rewrite the integrand:

      (1−cos⁡2(x))sin⁡(x)cos⁡4(x)=−sin⁡(x)cos⁡2(x)+sin⁡(x)cos⁡4(x)\frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos^{4}{\left(x \right)}} = \frac{- \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}}{\cos^{4}{\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 dudu:

      ∫u2−1u4 du\int \frac{u^{2} - 1}{u^{4}}\, du

      1. Rewrite the integrand:

        u2−1u4=1u2−1u4\frac{u^{2} - 1}{u^{4}} = \frac{1}{u^{2}} - \frac{1}{u^{4}}

      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:

          ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

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

          ∫(−1u4) du=−∫1u4 du\int \left(- \frac{1}{u^{4}}\right)\, du = - \int \frac{1}{u^{4}}\, du

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

            ∫1u4 du=−13u3\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}

          So, the result is: 13u3\frac{1}{3 u^{3}}

        The result is: −1u+13u3- \frac{1}{u} + \frac{1}{3 u^{3}}

      Now substitute uu back in:

      −1cos⁡(x)+13cos⁡3(x)- \frac{1}{\cos{\left(x \right)}} + \frac{1}{3 \cos^{3}{\left(x \right)}}

    Method #3

    1. Rewrite the integrand:

      (1−cos⁡2(x))sin⁡(x)cos⁡4(x)=−sin⁡(x)cos⁡2(x)+sin⁡(x)cos⁡4(x)\frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos^{4}{\left(x \right)}} = - \frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + \frac{\sin{\left(x \right)}}{\cos^{4}{\left(x \right)}}

    2. Integrate term-by-term:

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

        ∫(−sin⁡(x)cos⁡2(x)) dx=−∫sin⁡(x)cos⁡2(x) dx\int \left(- \frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}}\right)\, dx = - \int \frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}}\, dx

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

          Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

          ∫1u2 du\int \frac{1}{u^{2}}\, du

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

            ∫(−1u2) du=−∫1u2 du\int \left(- \frac{1}{u^{2}}\right)\, du = - \int \frac{1}{u^{2}}\, du

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

              ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

            So, the result is: 1u\frac{1}{u}

          Now substitute uu back in:

          1cos⁡(x)\frac{1}{\cos{\left(x \right)}}

        So, the result is: −1cos⁡(x)- \frac{1}{\cos{\left(x \right)}}

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

        Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

        ∫1u4 du\int \frac{1}{u^{4}}\, du

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

          ∫(−1u4) du=−∫1u4 du\int \left(- \frac{1}{u^{4}}\right)\, du = - \int \frac{1}{u^{4}}\, du

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

            ∫1u4 du=−13u3\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}

          So, the result is: 13u3\frac{1}{3 u^{3}}

        Now substitute uu back in:

        13cos⁡3(x)\frac{1}{3 \cos^{3}{\left(x \right)}}

      The result is: −1cos⁡(x)+13cos⁡3(x)- \frac{1}{\cos{\left(x \right)}} + \frac{1}{3 \cos^{3}{\left(x \right)}}

  3. Now simplify:

    −3+1cos⁡2(x)3cos⁡(x)\frac{-3 + \frac{1}{\cos^{2}{\left(x \right)}}}{3 \cos{\left(x \right)}}

  4. Add the constant of integration:

    −3+1cos⁡2(x)3cos⁡(x)+constant\frac{-3 + \frac{1}{\cos^{2}{\left(x \right)}}}{3 \cos{\left(x \right)}}+ \mathrm{constant}

The answer is:

−3+1cos⁡2(x)3cos⁡(x)+constant\frac{-3 + \frac{1}{\cos^{2}{\left(x \right)}}}{3 \cos{\left(x \right)}}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                   
 |                                    
 |    3                               
 | sin (x)            1          1    
 | ------- dx = C - ------ + ---------
 |    4             cos(x)        3   
 | cos (x)                   3*cos (x)
 |                                    
/
−3 cos⁡2x−13 cos⁡3x-{{3\,\cos ^2x-1}\over{3\,\cos ^3x}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-1010
Plot the graph f(x)
The answer [src] 
             2   
2   1 - 3*cos (1)
- + -------------
3          3     
      3*cos (1)
−1cos⁡1+13 cos⁡31+23-{{1}\over{\cos 1}}+{{1}\over{3\,\cos ^31}}+{{2}\over{3}} 
=
= 
             2   
2   1 - 3*cos (1)
- + -------------
3          3     
      3*cos (1)
−3cos⁡2(1)+13cos⁡3(1)+23\frac{- 3 \cos^{2}{\left(1 \right)} + 1}{3 \cos^{3}{\left(1 \right)}} + \frac{2}{3}
Simplify
Numerical answer [src] 
0.92918564057767
0.92918564057767
The graph
Integral of sin^3x/cos^4x dx

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

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