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sin^3x/cos^7x
Solving integrals/ sin^3x/cos^7x

Integral of sin^3x/cos^7x dx

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  1           
  /           
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 |     3      
 |  sin (x)   
 |  ------- dx
 |     7      
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0
∫01sin⁡3(x)cos⁡7(x) dx\int\limits_{0}^{1} \frac{\sin^{3}{\left(x \right)}}{\cos^{7}{\left(x \right)}}\, dx
Simplify
Detail solution

  1. Rewrite the integrand:

    sin⁡3(x)cos⁡7(x)=(1−cos⁡2(x))sin⁡(x)cos⁡7(x)\frac{\sin^{3}{\left(x \right)}}{\cos^{7}{\left(x \right)}} = \frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos^{7}{\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:

      ∫u2−1u7 du\int \frac{u^{2} - 1}{u^{7}}\, du

      1. Let u=u2u = u^{2}.

        Then let du=2ududu = 2 u du and substitute du2\frac{du}{2}:

        ∫u−14u4 du\int \frac{u - 1}{4 u^{4}}\, du

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

          ∫u−12u4 du=∫u−1u4 du2\int \frac{u - 1}{2 u^{4}}\, du = \frac{\int \frac{u - 1}{u^{4}}\, du}{2}

          1. Rewrite the integrand:

            u−1u4=1u3−1u4\frac{u - 1}{u^{4}} = \frac{1}{u^{3}} - \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:

              ∫1u3 du=−12u2\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}

            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: −12u2+13u3- \frac{1}{2 u^{2}} + \frac{1}{3 u^{3}}

          So, the result is: −14u2+16u3- \frac{1}{4 u^{2}} + \frac{1}{6 u^{3}}

        Now substitute uu back in:

        −14u4+16u6- \frac{1}{4 u^{4}} + \frac{1}{6 u^{6}}

      Now substitute uu back in:

      −14cos⁡4(x)+16cos⁡6(x)- \frac{1}{4 \cos^{4}{\left(x \right)}} + \frac{1}{6 \cos^{6}{\left(x \right)}}

    Method #2

    1. Rewrite the integrand:

      (1−cos⁡2(x))sin⁡(x)cos⁡7(x)=−sin⁡(x)cos⁡2(x)+sin⁡(x)cos⁡7(x)\frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos^{7}{\left(x \right)}} = \frac{- \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}}{\cos^{7}{\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:

      ∫u2−1u7 du\int \frac{u^{2} - 1}{u^{7}}\, du

      1. Let u=u2u = u^{2}.

        Then let du=2ududu = 2 u du and substitute du2\frac{du}{2}:

        ∫u−14u4 du\int \frac{u - 1}{4 u^{4}}\, du

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

          ∫u−12u4 du=∫u−1u4 du2\int \frac{u - 1}{2 u^{4}}\, du = \frac{\int \frac{u - 1}{u^{4}}\, du}{2}

          1. Rewrite the integrand:

            u−1u4=1u3−1u4\frac{u - 1}{u^{4}} = \frac{1}{u^{3}} - \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:

              ∫1u3 du=−12u2\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}

            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: −12u2+13u3- \frac{1}{2 u^{2}} + \frac{1}{3 u^{3}}

          So, the result is: −14u2+16u3- \frac{1}{4 u^{2}} + \frac{1}{6 u^{3}}

        Now substitute uu back in:

        −14u4+16u6- \frac{1}{4 u^{4}} + \frac{1}{6 u^{6}}

      Now substitute uu back in:

      −14cos⁡4(x)+16cos⁡6(x)- \frac{1}{4 \cos^{4}{\left(x \right)}} + \frac{1}{6 \cos^{6}{\left(x \right)}}

    Method #3

    1. Rewrite the integrand:

      (1−cos⁡2(x))sin⁡(x)cos⁡7(x)=−sin⁡(x)cos⁡5(x)+sin⁡(x)cos⁡7(x)\frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos^{7}{\left(x \right)}} = - \frac{\sin{\left(x \right)}}{\cos^{5}{\left(x \right)}} + \frac{\sin{\left(x \right)}}{\cos^{7}{\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⁡5(x)) dx=−∫sin⁡(x)cos⁡5(x) dx\int \left(- \frac{\sin{\left(x \right)}}{\cos^{5}{\left(x \right)}}\right)\, dx = - \int \frac{\sin{\left(x \right)}}{\cos^{5}{\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:

          ∫1u5 du\int \frac{1}{u^{5}}\, du

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

            ∫(−1u5) du=−∫1u5 du\int \left(- \frac{1}{u^{5}}\right)\, du = - \int \frac{1}{u^{5}}\, du

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

              ∫1u5 du=−14u4\int \frac{1}{u^{5}}\, du = - \frac{1}{4 u^{4}}

            So, the result is: 14u4\frac{1}{4 u^{4}}

          Now substitute uu back in:

          14cos⁡4(x)\frac{1}{4 \cos^{4}{\left(x \right)}}

        So, the result is: −14cos⁡4(x)- \frac{1}{4 \cos^{4}{\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:

        ∫1u7 du\int \frac{1}{u^{7}}\, du

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

          ∫(−1u7) du=−∫1u7 du\int \left(- \frac{1}{u^{7}}\right)\, du = - \int \frac{1}{u^{7}}\, du

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

            ∫1u7 du=−16u6\int \frac{1}{u^{7}}\, du = - \frac{1}{6 u^{6}}

          So, the result is: 16u6\frac{1}{6 u^{6}}

        Now substitute uu back in:

        16cos⁡6(x)\frac{1}{6 \cos^{6}{\left(x \right)}}

      The result is: −14cos⁡4(x)+16cos⁡6(x)- \frac{1}{4 \cos^{4}{\left(x \right)}} + \frac{1}{6 \cos^{6}{\left(x \right)}}

  3. Now simplify:

    2−3cos⁡2(x)12cos⁡6(x)\frac{2 - 3 \cos^{2}{\left(x \right)}}{12 \cos^{6}{\left(x \right)}}

  4. Add the constant of integration:

    2−3cos⁡2(x)12cos⁡6(x)+constant\frac{2 - 3 \cos^{2}{\left(x \right)}}{12 \cos^{6}{\left(x \right)}}+ \mathrm{constant}

The answer is:

2−3cos⁡2(x)12cos⁡6(x)+constant\frac{2 - 3 \cos^{2}{\left(x \right)}}{12 \cos^{6}{\left(x \right)}}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                      
 |                                       
 |    3                                  
 | sin (x)              1           1    
 | ------- dx = C - --------- + ---------
 |    7                  4           6   
 | cos (x)          4*cos (x)   6*cos (x)
 |                                       
/
−3 sin⁡2x−112 sin⁡6x−36 sin⁡4x+36 sin⁡2x−12-{{3\,\sin ^2x-1}\over{12\,\sin ^6x-36\,\sin ^4x+36\,\sin ^2x-12}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-5050
Plot the graph f(x)
The answer [src] 
              2   
1    2 - 3*cos (1)
-- + -------------
12           6    
       12*cos (1)
112 sin⁡61−36 sin⁡41+36 sin⁡21−12−sin⁡214 sin⁡61−12 sin⁡41+12 sin⁡21−4+112{{1}\over{12\,\sin ^61-36\,\sin ^41+36\,\sin ^21-12}}-{{\sin ^21 }\over{4\,\sin ^61-12\,\sin ^41+12\,\sin ^21-4}}+{{1}\over{12}} 
=
= 
              2   
1    2 - 3*cos (1)
-- + -------------
12           6    
       12*cos (1)
112+−3cos⁡2(1)+212cos⁡6(1)\frac{1}{12} + \frac{- 3 \cos^{2}{\left(1 \right)} + 2}{12 \cos^{6}{\left(1 \right)}}
Simplify
Numerical answer [src] 
3.84906381342323
3.84906381342323
The graph
Integral of sin^3x/cos^7x dx

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

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