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Solving integrals/ cos3xsin2xdx

Integral of cos3xsin2xdx dx

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∫0π2sin⁡(2x)cos⁡(3x) dx\int\limits_{0}^{\frac{\pi}{2}} \sin{\left(2 x \right)} \cos{\left(3 x \right)}\, dx 
Integral(cos(3*x)*sin(2*x), (x, 0, pi/2))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

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

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

      1. Rewrite the integrand:

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

          1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

            Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

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

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

              ∫u4 du=−∫u4 du\int u^{4}\, 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}

            Now substitute uu back in:

            −cos⁡5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

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

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

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

          1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

            Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

            ∫(−u2) du\int \left(- u^{2}\right)\, du

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

              ∫u2 du=−∫u2 du\int u^{2}\, 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}

              So, the result is: −u33- \frac{u^{3}}{3}

            Now substitute uu back in:

            −cos⁡3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

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

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

      So, the result is: −8cos⁡5(x)5+2cos⁡3(x)- \frac{8 \cos^{5}{\left(x \right)}}{5} + 2 \cos^{3}{\left(x \right)}

    Method #2

    1. Rewrite the integrand:

      sin⁡(2x)cos⁡(3x)=8sin⁡(x)cos⁡4(x)−6sin⁡(x)cos⁡2(x)\sin{\left(2 x \right)} \cos{\left(3 x \right)} = 8 \sin{\left(x \right)} \cos^{4}{\left(x \right)} - 6 \sin{\left(x \right)} \cos^{2}{\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:

        ∫8sin⁡(x)cos⁡4(x) dx=8∫sin⁡(x)cos⁡4(x) dx\int 8 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = 8 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx

        1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

          Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

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

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

            ∫u4 du=−∫u4 du\int u^{4}\, 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}

          Now substitute uu back in:

          −cos⁡5(x)5- \frac{\cos^{5}{\left(x \right)}}{5}

        So, the result is: −8cos⁡5(x)5- \frac{8 \cos^{5}{\left(x \right)}}{5}

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

        ∫(−6sin⁡(x)cos⁡2(x)) dx=−6∫sin⁡(x)cos⁡2(x) dx\int \left(- 6 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 6 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx

        1. Let u=cos⁡(x)u = \cos{\left(x \right)}.

          Then let du=−sin⁡(x)dxdu = - \sin{\left(x \right)} dx and substitute −du- du:

          ∫(−u2) du\int \left(- u^{2}\right)\, du

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

            ∫u2 du=−∫u2 du\int u^{2}\, 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}

            So, the result is: −u33- \frac{u^{3}}{3}

          Now substitute uu back in:

          −cos⁡3(x)3- \frac{\cos^{3}{\left(x \right)}}{3}

        So, the result is: 2cos⁡3(x)2 \cos^{3}{\left(x \right)}

      The result is: −8cos⁡5(x)5+2cos⁡3(x)- \frac{8 \cos^{5}{\left(x \right)}}{5} + 2 \cos^{3}{\left(x \right)}

  2. Now simplify:

    cos⁡(x)2−cos⁡(5x)10\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(5 x \right)}}{10}

  3. Add the constant of integration:

    cos⁡(x)2−cos⁡(5x)10+constant\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(5 x \right)}}{10}+ \mathrm{constant}

The answer is:

cos⁡(x)2−cos⁡(5x)10+constant\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(5 x \right)}}{10}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                            5   
 |                                 3      8*cos (x)
 | cos(3*x)*sin(2*x) dx = C + 2*cos (x) - ---------
 |                                            5    
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∫sin⁡(2x)cos⁡(3x) dx=C−8cos⁡5(x)5+2cos⁡3(x)\int \sin{\left(2 x \right)} \cos{\left(3 x \right)}\, dx = C - \frac{8 \cos^{5}{\left(x \right)}}{5} + 2 \cos^{3}{\left(x \right)}
Simplify
The graph
0.00.10.20.30.40.50.60.70.80.91.01.11.21.31.41.52-2
Plot the graph f(x)
The answer [src] 
-2/5
−25- \frac{2}{5} 
=
= 
-2/5
−25- \frac{2}{5} 
-2/5
Numerical answer [src] 
-0.4
-0.4

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

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