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Solving integrals/ (2sin5x+3cosx/2)

Integral of (2sin5x+3cosx/2) dx

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The solution

You have entered [src] 
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 |  /             3*cos(x)\   
 |  |2*sin(5*x) + --------| dx
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∫01(2sin⁡(5x)+3cos⁡(x)2) dx\int\limits_{0}^{1} \left(2 \sin{\left(5 x \right)} + \frac{3 \cos{\left(x \right)}}{2}\right)\, dx 
Integral(2*sin(5*x) + (3*cos(x))/2, (x, 0, 1))
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:

      ∫2sin⁡(5x) dx=2∫sin⁡(5x) dx\int 2 \sin{\left(5 x \right)}\, dx = 2 \int \sin{\left(5 x \right)}\, dx

      1. Let u=5xu = 5 x.

        Then let du=5dxdu = 5 dx and substitute du5\frac{du}{5}:

        ∫sin⁡(u)5 du\int \frac{\sin{\left(u \right)}}{5}\, du

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

          ∫sin⁡(u) du=∫sin⁡(u) du5\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}

          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)5- \frac{\cos{\left(u \right)}}{5}

        Now substitute uu back in:

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

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

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

      ∫3cos⁡(x)2 dx=∫3cos⁡(x) dx2\int \frac{3 \cos{\left(x \right)}}{2}\, dx = \frac{\int 3 \cos{\left(x \right)}\, dx}{2}

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

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

        1. The integral of cosine is sine:

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

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

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

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

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                                      
 |                                                       
 | /             3*cos(x)\          2*cos(5*x)   3*sin(x)
 | |2*sin(5*x) + --------| dx = C - ---------- + --------
 | \                2    /              5           2    
 |                                                       
/
∫(2sin⁡(5x)+3cos⁡(x)2) dx=C+3sin⁡(x)2−2cos⁡(5x)5\int \left(2 \sin{\left(5 x \right)} + \frac{3 \cos{\left(x \right)}}{2}\right)\, dx = C + \frac{3 \sin{\left(x \right)}}{2} - \frac{2 \cos{\left(5 x \right)}}{5}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
Plot the graph f(x)
The answer [src] 
2   2*cos(5)   3*sin(1)
- - -------- + --------
5      5          2
−2cos⁡(5)5+25+3sin⁡(1)2- \frac{2 \cos{\left(5 \right)}}{5} + \frac{2}{5} + \frac{3 \sin{\left(1 \right)}}{2} 
=
= 
2   2*cos(5)   3*sin(1)
- - -------- + --------
5      5          2
−2cos⁡(5)5+25+3sin⁡(1)2- \frac{2 \cos{\left(5 \right)}}{5} + \frac{2}{5} + \frac{3 \sin{\left(1 \right)}}{2} 
2/5 - 2*cos(5)/5 + 3*sin(1)/2
Numerical answer [src] 
1.54874160302655
1.54874160302655

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

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