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cos^5x/sin^2x
Solving integrals/ cos^5x/sin^2x

Integral of cos^5x/sin^2x dx

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∫01cos⁡5(x)sin⁡2(x) dx\int\limits_{0}^{1} \frac{\cos^{5}{\left(x \right)}}{\sin^{2}{\left(x \right)}}\, dx 
Integral(cos(x)^5/(sin(x)^2), (x, 0, 1))
Detail solution

  1. Rewrite the integrand:

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

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

      1. Rewrite the integrand:

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

      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:

          ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

        1. The integral of a constant is the constant times the variable of integration:

          ∫(−2) du=−2u\int \left(-2\right)\, du = - 2 u

        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}

        The result is: u33−2u−1u\frac{u^{3}}{3} - 2 u - \frac{1}{u}

      Now substitute uu back in:

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

    Method #2

    1. Rewrite the integrand:

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

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

      1. Rewrite the integrand:

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

      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:

          ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

        1. The integral of a constant is the constant times the variable of integration:

          ∫(−2) du=−2u\int \left(-2\right)\, du = - 2 u

        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}

        The result is: u33−2u−1u\frac{u^{3}}{3} - 2 u - \frac{1}{u}

      Now substitute uu back in:

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

    Method #3

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

      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 du\int u^{2}\, du

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

          ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

        Now substitute uu back in:

        sin⁡3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

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

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

        1. The integral of cosine is sine:

          ∫cos⁡(x) dx=sin⁡(x)\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}

        So, the result is: −2sin⁡(x)- 2 \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:

        ∫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)}}

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

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                            
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 |    5                                    3   
 | cos (x)            1                 sin (x)
 | ------- dx = C - ------ - 2*sin(x) + -------
 |    2             sin(x)                 3   
 | sin (x)                                     
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/
sin⁡3x−6 sin⁡x3−1sin⁡x{{\sin ^3x-6\,\sin x}\over{3}}-{{1}\over{\sin x}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-100000000100000000
Plot the graph f(x)
The answer [src] 
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Simplify
Numerical answer [src] 
1.3793236779486e+19
1.3793236779486e+19
The graph
Integral of cos^5x/sin^2x dx

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

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