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(cos^3x/sin^6x)
Solving integrals/ (cos^3x/sin^6x)

Integral of (cos^3x/sin^6x) dx

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∫01cos⁡3(x)sin⁡6(x) dx\int\limits_{0}^{1} \frac{\cos^{3}{\left(x \right)}}{\sin^{6}{\left(x \right)}}\, dx 
Integral(cos(x)^3/(sin(x)^6), (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

    cos⁡3(x)sin⁡6(x)=(1−sin⁡2(x))cos⁡(x)sin⁡6(x)\frac{\cos^{3}{\left(x \right)}}{\sin^{6}{\left(x \right)}} = \frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{6}{\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 dudu:

      ∫1−u2u6 du\int \frac{1 - u^{2}}{u^{6}}\, du

      1. Rewrite the integrand:

        1−u2u6=−1u4+1u6\frac{1 - u^{2}}{u^{6}} = - \frac{1}{u^{4}} + \frac{1}{u^{6}}

      2. Integrate term-by-term:

        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}}

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

          ∫1u6 du=−15u5\int \frac{1}{u^{6}}\, du = - \frac{1}{5 u^{5}}

        The result is: 13u3−15u5\frac{1}{3 u^{3}} - \frac{1}{5 u^{5}}

      Now substitute uu back in:

      13sin⁡3(x)−15sin⁡5(x)\frac{1}{3 \sin^{3}{\left(x \right)}} - \frac{1}{5 \sin^{5}{\left(x \right)}}

    Method #2

    1. Rewrite the integrand:

      (1−sin⁡2(x))cos⁡(x)sin⁡6(x)=−sin⁡2(x)cos⁡(x)+cos⁡(x)sin⁡6(x)\frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{6}{\left(x \right)}} = \frac{- \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}}{\sin^{6}{\left(x \right)}}

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

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

      ∫1−u2u6 du\int \frac{1 - u^{2}}{u^{6}}\, du

      1. Rewrite the integrand:

        1−u2u6=−1u4+1u6\frac{1 - u^{2}}{u^{6}} = - \frac{1}{u^{4}} + \frac{1}{u^{6}}

      2. Integrate term-by-term:

        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}}

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

          ∫1u6 du=−15u5\int \frac{1}{u^{6}}\, du = - \frac{1}{5 u^{5}}

        The result is: 13u3−15u5\frac{1}{3 u^{3}} - \frac{1}{5 u^{5}}

      Now substitute uu back in:

      13sin⁡3(x)−15sin⁡5(x)\frac{1}{3 \sin^{3}{\left(x \right)}} - \frac{1}{5 \sin^{5}{\left(x \right)}}

    Method #3

    1. Rewrite the integrand:

      (1−sin⁡2(x))cos⁡(x)sin⁡6(x)=−cos⁡(x)sin⁡4(x)+cos⁡(x)sin⁡6(x)\frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{6}{\left(x \right)}} = - \frac{\cos{\left(x \right)}}{\sin^{4}{\left(x \right)}} + \frac{\cos{\left(x \right)}}{\sin^{6}{\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⁡4(x)) dx=−∫cos⁡(x)sin⁡4(x) dx\int \left(- \frac{\cos{\left(x \right)}}{\sin^{4}{\left(x \right)}}\right)\, dx = - \int \frac{\cos{\left(x \right)}}{\sin^{4}{\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:

          ∫1u4 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}}

          Now substitute uu back in:

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

        So, the result is: 13sin⁡3(x)\frac{1}{3 \sin^{3}{\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:

        ∫1u6 du\int \frac{1}{u^{6}}\, du

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

          ∫1u6 du=−15u5\int \frac{1}{u^{6}}\, du = - \frac{1}{5 u^{5}}

        Now substitute uu back in:

        −15sin⁡5(x)- \frac{1}{5 \sin^{5}{\left(x \right)}}

      The result is: 13sin⁡3(x)−15sin⁡5(x)\frac{1}{3 \sin^{3}{\left(x \right)}} - \frac{1}{5 \sin^{5}{\left(x \right)}}

  3. Now simplify:

    5sin⁡2(x)−315sin⁡5(x)\frac{5 \sin^{2}{\left(x \right)} - 3}{15 \sin^{5}{\left(x \right)}}

  4. Add the constant of integration:

    5sin⁡2(x)−315sin⁡5(x)+constant\frac{5 \sin^{2}{\left(x \right)} - 3}{15 \sin^{5}{\left(x \right)}}+ \mathrm{constant}

The answer is:

5sin⁡2(x)−315sin⁡5(x)+constant\frac{5 \sin^{2}{\left(x \right)} - 3}{15 \sin^{5}{\left(x \right)}}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                      
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 |    3                                  
 | cos (x)              1           1    
 | ------- dx = C - --------- + ---------
 |    6                  5           3   
 | sin (x)          5*sin (x)   3*sin (x)
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∫cos⁡3(x)sin⁡6(x) dx=C+13sin⁡3(x)−15sin⁡5(x)\int \frac{\cos^{3}{\left(x \right)}}{\sin^{6}{\left(x \right)}}\, dx = C + \frac{1}{3 \sin^{3}{\left(x \right)}} - \frac{1}{5 \sin^{5}{\left(x \right)}}
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The graph
0.001.000.100.200.300.400.500.600.700.800.90-1e231e23
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The answer [src] 
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Numerical answer [src] 
7.0110751903966e+94
7.0110751903966e+94
The graph
Integral of (cos^3x/sin^6x) dx

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

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