Mister Exam

Other calculators

cos3x(sinx)^2
Solving integrals/ cos3x(sinx)^2

Integral of cos3x(sinx)^2 dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  3                    
  /                    
 |                     
 |              2      
 |  cos(3*x)*sin (x) dx
 |                     
/                      
1
∫13sin⁡2(x)cos⁡(3x) dx\int\limits_{1}^{3} \sin^{2}{\left(x \right)} \cos{\left(3 x \right)}\, dx 
Integral(cos(3*x)*sin(x)^2, (x, 1, 3))
Detail solution

  1. Rewrite the integrand:

    sin⁡2(x)cos⁡(3x)=4sin⁡2(x)cos⁡3(x)−3sin⁡2(x)cos⁡(x)\sin^{2}{\left(x \right)} \cos{\left(3 x \right)} = 4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)} - 3 \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:

      ∫4sin⁡2(x)cos⁡3(x) dx=4∫sin⁡2(x)cos⁡3(x) dx\int 4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = 4 \int \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx

      1. Rewrite the integrand:

        sin⁡2(x)cos⁡3(x)=(1−sin⁡2(x))sin⁡2(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=−∫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}

      So, the result is: −4sin⁡5(x)5+4sin⁡3(x)3- \frac{4 \sin^{5}{\left(x \right)}}{5} + \frac{4 \sin^{3}{\left(x \right)}}{3}

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

      ∫(−3sin⁡2(x)cos⁡(x)) dx=−3∫sin⁡2(x)cos⁡(x) dx\int \left(- 3 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 3 \int \sin^{2}{\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:

        ∫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}

      So, the result is: −sin⁡3(x)- \sin^{3}{\left(x \right)}

    The result is: −4sin⁡5(x)5+sin⁡3(x)3- \frac{4 \sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}

  3. Now simplify:

    (5−12sin⁡2(x))sin⁡3(x)15\frac{\left(5 - 12 \sin^{2}{\left(x \right)}\right) \sin^{3}{\left(x \right)}}{15}

  4. Add the constant of integration:

    (5−12sin⁡2(x))sin⁡3(x)15+constant\frac{\left(5 - 12 \sin^{2}{\left(x \right)}\right) \sin^{3}{\left(x \right)}}{15}+ \mathrm{constant}

The answer is:

(5−12sin⁡2(x))sin⁡3(x)15+constant\frac{\left(5 - 12 \sin^{2}{\left(x \right)}\right) \sin^{3}{\left(x \right)}}{15}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                             
 |                                5         3   
 |             2             4*sin (x)   sin (x)
 | cos(3*x)*sin (x) dx = C - --------- + -------
 |                               5          3   
/
−3 sin⁡5x−5 sin⁡3x15−3 sin⁡5x5-{{3\,\sin ^5x-5\,\sin ^3x}\over{15}}-{{3\,\sin ^5x}\over{5}}
Simplify
The graph
1.03.01.21.41.61.82.02.22.42.62.82-2
Plot the graph f(x)
The answer [src] 
       2                  2                  2                  2                                                            
  7*sin (1)*sin(3)   2*cos (3)*sin(9)   2*cos (1)*sin(3)   7*sin (3)*sin(9)   2*cos(1)*cos(3)*sin(1)   2*cos(3)*cos(9)*sin(3)
- ---------------- - ---------------- + ---------------- + ---------------- - ---------------------- + ----------------------
         15                 15                 15                 15                    5                        5
12 sin⁡51−5 sin⁡3115−12 sin⁡53−5 sin⁡3315{{12\,\sin ^51-5\,\sin ^31}\over{15}}-{{12\,\sin ^53-5\,\sin ^33 }\over{15}} 
=
= 
       2                  2                  2                  2                                                            
  7*sin (1)*sin(3)   2*cos (3)*sin(9)   2*cos (1)*sin(3)   7*sin (3)*sin(9)   2*cos(1)*cos(3)*sin(1)   2*cos(3)*cos(9)*sin(3)
- ---------------- - ---------------- + ---------------- + ---------------- - ---------------------- + ----------------------
         15                 15                 15                 15                    5                        5
−2sin⁡(9)cos⁡2(3)15−7sin⁡2(1)sin⁡(3)15+7sin⁡2(3)sin⁡(9)15+2sin⁡(3)cos⁡2(1)15+2sin⁡(3)cos⁡(3)cos⁡(9)5−2sin⁡(1)cos⁡(1)cos⁡(3)5- \frac{2 \sin{\left(9 \right)} \cos^{2}{\left(3 \right)}}{15} - \frac{7 \sin^{2}{\left(1 \right)} \sin{\left(3 \right)}}{15} + \frac{7 \sin^{2}{\left(3 \right)} \sin{\left(9 \right)}}{15} + \frac{2 \sin{\left(3 \right)} \cos^{2}{\left(1 \right)}}{15} + \frac{2 \sin{\left(3 \right)} \cos{\left(3 \right)} \cos{\left(9 \right)}}{5} - \frac{2 \sin{\left(1 \right)} \cos{\left(1 \right)} \cos{\left(3 \right)}}{5}
Simplify
Numerical answer [src] 
0.139793551309643
0.139793551309643
The graph
Integral of cos3x(sinx)^2 dx

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

    © Mister Exam – Calculator