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Solving integrals/ 2sin2x+12/(5pi)x

Integral of 2sin2x+12/(5pi)x dx

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The solution

You have entered [src] 
 5*pi                        
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  12                         
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  |  /              12   \   
  |  |2*sin(2*x) + ----*x| dx
  |  \             5*pi  /   
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 0
∫05π12(x125π+2sin⁡(2x)) dx\int\limits_{0}^{\frac{5 \pi}{12}} \left(x \frac{12}{5 \pi} + 2 \sin{\left(2 x \right)}\right)\, dx 
Integral(2*sin(2*x) + (12/((5*pi)))*x, (x, 0, 5*pi/12))
Detail solution

  1. Integrate term-by-term:

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

      ∫x125π dx=1215π∫x dx\int x \frac{12}{5 \pi}\, dx = 12 \frac{1}{5 \pi} \int x\, dx

      1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

      So, the result is: 615πx26 \frac{1}{5 \pi} x^{2}

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

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

      1. There are multiple ways to do this integral.

        Method #1

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

          ∫sin⁡(u)2 du\int \frac{\sin{\left(u \right)}}{2}\, du

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

            ∫sin⁡(u) du=∫sin⁡(u) du2\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}

            1. The integral of sine is negative cosine:

              ∫sin⁡(u) du=−cos⁡(u)\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}

            So, the result is: −cos⁡(u)2- \frac{\cos{\left(u \right)}}{2}

          Now substitute uu back in:

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

        Method #2

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

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

          1. There are multiple ways to do this integral.

            Method #1

            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:

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

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

                ∫u du=−∫u du\int u\, du = - \int u\, du

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

                  ∫u du=u22\int u\, du = \frac{u^{2}}{2}

                So, the result is: −u22- \frac{u^{2}}{2}

              Now substitute uu back in:

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

            Method #2

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

              Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

              ∫u du\int u\, du

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

                ∫u du=u22\int u\, du = \frac{u^{2}}{2}

              Now substitute uu back in:

              sin⁡2(x)2\frac{\sin^{2}{\left(x \right)}}{2}

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

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

    The result is: 615πx2−cos⁡(2x)6 \frac{1}{5 \pi} x^{2} - \cos{\left(2 x \right)}

  2. Now simplify:

    6x25π−cos⁡(2x)\frac{6 x^{2}}{5 \pi} - \cos{\left(2 x \right)}

  3. Add the constant of integration:

    6x25π−cos⁡(2x)+constant\frac{6 x^{2}}{5 \pi} - \cos{\left(2 x \right)}+ \mathrm{constant}

The answer is:

6x25π−cos⁡(2x)+constant\frac{6 x^{2}}{5 \pi} - \cos{\left(2 x \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                   
 |                                                    
 | /              12   \                        2  1  
 | |2*sin(2*x) + ----*x| dx = C - cos(2*x) + 6*x *----
 | \             5*pi  /                          5*pi
 |                                                    
/
∫(x125π+2sin⁡(2x)) dx=C+615πx2−cos⁡(2x)\int \left(x \frac{12}{5 \pi} + 2 \sin{\left(2 x \right)}\right)\, dx = C + 6 \frac{1}{5 \pi} x^{2} - \cos{\left(2 x \right)}
Simplify
The graph
0.00.10.20.30.40.50.60.70.80.91.01.11.21.35-5
Plot the graph f(x)
The answer [src] 
      ___       
    \/ 3    5*pi
1 + ----- + ----
      2      24
5π24+32+1\frac{5 \pi}{24} + \frac{\sqrt{3}}{2} + 1 
=
= 
      ___       
    \/ 3    5*pi
1 + ----- + ----
      2      24
5π24+32+1\frac{5 \pi}{24} + \frac{\sqrt{3}}{2} + 1 
1 + sqrt(3)/2 + 5*pi/24
Numerical answer [src] 
2.52052387328231
2.52052387328231

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

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