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dx/sin^5xcos^3x
Solving integrals/ dx/sin^5xcos^3x

Integral of dx/sin^5xcos^3x dx

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∫01cos⁡3(x)sin⁡5(x) dx\int\limits_{0}^{1} \frac{\cos^{3}{\left(x \right)}}{\sin^{5}{\left(x \right)}}\, dx 
Integral(cos(x)^3/sin(x)^5, (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    cos⁡3(x)sin⁡5(x)=(1−sin⁡2(x))cos⁡(x)sin⁡5(x)\frac{\cos^{3}{\left(x \right)}}{\sin^{5}{\left(x \right)}} = \frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{5}{\left(x \right)}}

  2. There are multiple ways to do this integral.

    Method #1

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

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

      ∫(−u2−1u5) du\int \left(- \frac{u^{2} - 1}{u^{5}}\right)\, du

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

        ∫u2−1u5 du=−∫u2−1u5 du\int \frac{u^{2} - 1}{u^{5}}\, du = - \int \frac{u^{2} - 1}{u^{5}}\, du

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

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

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

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

            ∫u−1u3 du=∫u−1u3 du2\int \frac{u - 1}{u^{3}}\, du = \frac{\int \frac{u - 1}{u^{3}}\, du}{2}

            1. Rewrite the integrand:

              u−1u3=1u2−1u3\frac{u - 1}{u^{3}} = \frac{1}{u^{2}} - \frac{1}{u^{3}}

            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:

                ∫(−1u3) du=−∫1u3 du\int \left(- \frac{1}{u^{3}}\right)\, du = - \int \frac{1}{u^{3}}\, du

                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}}

                So, the result is: 12u2\frac{1}{2 u^{2}}

              The result is: −1u+12u2- \frac{1}{u} + \frac{1}{2 u^{2}}

            So, the result is: −12u+14u2- \frac{1}{2 u} + \frac{1}{4 u^{2}}

          Now substitute uu back in:

          −12u2+14u4- \frac{1}{2 u^{2}} + \frac{1}{4 u^{4}}

        So, the result is: 12u2−14u4\frac{1}{2 u^{2}} - \frac{1}{4 u^{4}}

      Now substitute uu back in:

      12sin⁡2(x)−14sin⁡4(x)\frac{1}{2 \sin^{2}{\left(x \right)}} - \frac{1}{4 \sin^{4}{\left(x \right)}}

    Method #2

    1. Rewrite the integrand:

      (1−sin⁡2(x))cos⁡(x)sin⁡5(x)=−sin⁡2(x)cos⁡(x)−cos⁡(x)sin⁡5(x)\frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{5}{\left(x \right)}} = - \frac{\sin^{2}{\left(x \right)} \cos{\left(x \right)} - \cos{\left(x \right)}}{\sin^{5}{\left(x \right)}}

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

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

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

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

        ∫u2−1u5 du\int \frac{u^{2} - 1}{u^{5}}\, du

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

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

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

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

            ∫u−1u3 du=∫u−1u3 du2\int \frac{u - 1}{u^{3}}\, du = \frac{\int \frac{u - 1}{u^{3}}\, du}{2}

            1. Rewrite the integrand:

              u−1u3=1u2−1u3\frac{u - 1}{u^{3}} = \frac{1}{u^{2}} - \frac{1}{u^{3}}

            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:

                ∫(−1u3) du=−∫1u3 du\int \left(- \frac{1}{u^{3}}\right)\, du = - \int \frac{1}{u^{3}}\, du

                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}}

                So, the result is: 12u2\frac{1}{2 u^{2}}

              The result is: −1u+12u2- \frac{1}{u} + \frac{1}{2 u^{2}}

            So, the result is: −12u+14u2- \frac{1}{2 u} + \frac{1}{4 u^{2}}

          Now substitute uu back in:

          −12u2+14u4- \frac{1}{2 u^{2}} + \frac{1}{4 u^{4}}

        Now substitute uu back in:

        −12sin⁡2(x)+14sin⁡4(x)- \frac{1}{2 \sin^{2}{\left(x \right)}} + \frac{1}{4 \sin^{4}{\left(x \right)}}

      So, the result is: 12sin⁡2(x)−14sin⁡4(x)\frac{1}{2 \sin^{2}{\left(x \right)}} - \frac{1}{4 \sin^{4}{\left(x \right)}}

    Method #3

    1. Rewrite the integrand:

      (1−sin⁡2(x))cos⁡(x)sin⁡5(x)=−cos⁡(x)sin⁡3(x)+cos⁡(x)sin⁡5(x)\frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{5}{\left(x \right)}} = - \frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}} + \frac{\cos{\left(x \right)}}{\sin^{5}{\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:

        ∫(−cos⁡(x)sin⁡3(x)) dx=−∫cos⁡(x)sin⁡3(x) dx\int \left(- \frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\right)\, dx = - \int \frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx

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

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

          ∫1u3 du\int \frac{1}{u^{3}}\, du

          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}}

          Now substitute uu back in:

          −12sin⁡2(x)- \frac{1}{2 \sin^{2}{\left(x \right)}}

        So, the result is: 12sin⁡2(x)\frac{1}{2 \sin^{2}{\left(x \right)}}

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

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

        ∫1u5 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}}

        Now substitute uu back in:

        −14sin⁡4(x)- \frac{1}{4 \sin^{4}{\left(x \right)}}

      The result is: 12sin⁡2(x)−14sin⁡4(x)\frac{1}{2 \sin^{2}{\left(x \right)}} - \frac{1}{4 \sin^{4}{\left(x \right)}}

  3. Now simplify:

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

  4. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                      
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 |    3                                  
 | cos (x)              1           1    
 | ------- dx = C + --------- - ---------
 |    5                  2           4   
 | sin (x)          2*sin (x)   4*sin (x)
 |                                       
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∫cos⁡3(x)sin⁡5(x) dx=C+12sin⁡2(x)−14sin⁡4(x)\int \frac{\cos^{3}{\left(x \right)}}{\sin^{5}{\left(x \right)}}\, dx = C + \frac{1}{2 \sin^{2}{\left(x \right)}} - \frac{1}{4 \sin^{4}{\left(x \right)}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-5000000000000000000050000000000000000000
Plot the graph f(x)
The answer [src] 
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Numerical answer [src] 
7.26749061658134e+75
7.26749061658134e+75
The graph
Integral of dx/sin^5xcos^3x dx

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

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