Mister Exam

Integral of cos³xsin²x dx

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The solution

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01sin2(x)cos3(x)dx\int\limits_{0}^{1} \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx
Integral(cos(x)^3*sin(x)^2, (x, 0, 1))
Detail solution
  1. Rewrite the integrand:

    sin2(x)cos3(x)=(1sin2(x))sin2(x)cos(x)\sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)} = \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=u4du\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 n1n \neq -1:

            u4du=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 n1n \neq -1:

          u2du=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:

      sin5(x)5+sin3(x)3- \frac{\sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}

    Method #2

    1. Rewrite the integrand:

      (1sin2(x))sin2(x)cos(x)=sin4(x)cos(x)+sin2(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:

        (sin4(x)cos(x))dx=sin4(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:

          u4du\int u^{4}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

            u4du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          Now substitute uu back in:

          sin5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

        So, the result is: sin5(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:

        u2du\int u^{2}\, du

        1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

          u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

        Now substitute uu back in:

        sin3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

      The result is: sin5(x)5+sin3(x)3- \frac{\sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}

    Method #3

    1. Rewrite the integrand:

      (1sin2(x))sin2(x)cos(x)=sin4(x)cos(x)+sin2(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:

        (sin4(x)cos(x))dx=sin4(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:

          u4du\int u^{4}\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

            u4du=u55\int u^{4}\, du = \frac{u^{5}}{5}

          Now substitute uu back in:

          sin5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

        So, the result is: sin5(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:

        u2du\int u^{2}\, du

        1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

          u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

        Now substitute uu back in:

        sin3(x)3\frac{\sin^{3}{\left(x \right)}}{3}

      The result is: sin5(x)5+sin3(x)3- \frac{\sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}

  3. Add the constant of integration:

    sin5(x)5+sin3(x)3+constant- \frac{\sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}+ \mathrm{constant}


The answer is:

sin5(x)5+sin3(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) dx = C - ------- + -------
 |                             5         3   
/                                            
3sin5x5sin3x15-{{3\,\sin ^5x-5\,\sin ^3x}\over{15}}
The graph
0.001.000.100.200.300.400.500.600.700.800.900.00.2
The answer [src]
     5         3   
  sin (1)   sin (1)
- ------- + -------
     5         3   
3sin515sin3115-{{3\,\sin ^51-5\,\sin ^31}\over{15}}
=
=
     5         3   
  sin (1)   sin (1)
- ------- + -------
     5         3   
sin5(1)5+sin3(1)3- \frac{\sin^{5}{\left(1 \right)}}{5} + \frac{\sin^{3}{\left(1 \right)}}{3}
Numerical answer [src]
0.114230426366362
0.114230426366362
The graph
Integral of cos³xsin²x dx

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