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Solving integrals/ cos^3xsin^2xdx

Integral of cos^3xsin^2xdx dx

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∫0p2cos⁡3(x)sin⁡2(x)1 dx\int\limits_{0}^{\frac{p}{2}} \cos^{3}{\left(x \right)} \sin^{2}{\left(x \right)} 1\, dx 
Integral(cos(x)^3*sin(x)^2*1, (x, 0, p/2))
Detail solution

  1. Rewrite the integrand:

    cos⁡3(x)sin⁡2(x)1=(1−sin⁡2(x))sin⁡2(x)cos⁡(x)\cos^{3}{\left(x \right)} \sin^{2}{\left(x \right)} 1 = \left(1 - \sin^{2}{\left(x \right)}\right) \sin^{2}{\left(x \right)} \cos{\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+u2) du\int \left(- u^{4} + u^{2}\right)\, du

      1. Integrate term-by-term:

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

          ∫(−u4) du=−∫u4 du\int \left(- u^{4}\right)\, du = - \int u^{4}\, du

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

            ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          So, the result is: −u55- \frac{u^{5}}{5}

        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}

        The result is: −u55+u33- \frac{u^{5}}{5} + \frac{u^{3}}{3}

      Now substitute uu back in:

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

    Method #2

    1. Rewrite the integrand:

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

          ∫u4 du\int u^{4}\, du

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

            ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          Now substitute uu back in:

          sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

        So, the result is: −sin⁡5(x)5- \frac{\sin^{5}{\left(x \right)}}{5}

      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}

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

    Method #3

    1. Rewrite the integrand:

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

          ∫u4 du\int u^{4}\, du

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

            ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          Now substitute uu back in:

          sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

        So, the result is: −sin⁡5(x)5- \frac{\sin^{5}{\left(x \right)}}{5}

      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}

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

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                            
 |                               5         3   
 |    3       2               sin (x)   sin (x)
 | cos (x)*sin (x)*1 dx = C - ------- + -------
 |                               5         3   
/
−3 sin⁡5x−5 sin⁡3x15-{{3\,\sin ^5x-5\,\sin ^3x}\over{15}}
Simplify
The answer [src] 
     5/p\      3/p\
  sin |-|   sin |-|
      \2/       \2/
- ------- + -------
     5         3
−sin⁡5(p2)5+sin⁡3(p2)3- \frac{\sin^{5}{\left(\frac{p}{2} \right)}}{5} + \frac{\sin^{3}{\left(\frac{p}{2} \right)}}{3} 
=
= 
     5/p\      3/p\
  sin |-|   sin |-|
      \2/       \2/
- ------- + -------
     5         3
−sin⁡5(p2)5+sin⁡3(p2)3- \frac{\sin^{5}{\left(\frac{p}{2} \right)}}{5} + \frac{\sin^{3}{\left(\frac{p}{2} \right)}}{3}
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    Use the examples entering the upper and lower limits of integration.

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