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

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

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

  1. Rewrite the integrand:

    cos⁡3(x)sin⁡4(x)=(1−sin⁡2(x))cos⁡(x)sin⁡4(x)\frac{\cos^{3}{\left(x \right)}}{\sin^{4}{\left(x \right)}} = \frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{4}{\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−u2u4 du\int \frac{1 - u^{2}}{u^{4}}\, du

      1. Rewrite the integrand:

        1−u2u4=−1u2+1u4\frac{1 - u^{2}}{u^{4}} = - \frac{1}{u^{2}} + \frac{1}{u^{4}}

      2. Integrate term-by-term:

        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}

        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}}

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

      Now substitute uu back in:

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

    Method #2

    1. Rewrite the integrand:

      (1−sin⁡2(x))cos⁡(x)sin⁡4(x)=−sin⁡2(x)cos⁡(x)+cos⁡(x)sin⁡4(x)\frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{4}{\left(x \right)}} = \frac{- \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}}{\sin^{4}{\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−u2u4 du\int \frac{1 - u^{2}}{u^{4}}\, du

      1. Rewrite the integrand:

        1−u2u4=−1u2+1u4\frac{1 - u^{2}}{u^{4}} = - \frac{1}{u^{2}} + \frac{1}{u^{4}}

      2. Integrate term-by-term:

        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}

        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}}

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

      Now substitute uu back in:

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

    Method #3

    1. Rewrite the integrand:

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

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

          ∫1u2 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}

          Now substitute uu back in:

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

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

        ∫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)}}

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

  3. Now simplify:

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

  4. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                   
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 |    3                               
 | cos (x)            1          1    
 | ------- dx = C + ------ - ---------
 |    4             sin(x)        3   
 | sin (x)                   3*sin (x)
 |                                    
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3 sin⁡2x−13 sin⁡3x{{3\,\sin ^2x-1}\over{3\,\sin ^3x}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9020000000000000000-10000000000000000
Plot the graph f(x)
The answer [src] 
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Simplify
Numerical answer [src] 
7.81431122445857e+56
7.81431122445857e+56
The graph
Integral of cos^3x/sin^4x dx

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

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