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sin(2x+3)
Solving integrals/ sin(2x+3)

Integral of sin(2x+3) dx

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The solution

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∫01sin⁡(2x+3) dx\int\limits_{0}^{1} \sin{\left(2 x + 3 \right)}\, dx 
Integral(sin(2*x + 3), (x, 0, 1))
Detail solution

  1. Let u=2x+3u = 2 x + 3.

    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+3)2- \frac{\cos{\left(2 x + 3 \right)}}{2}

  2. Now simplify:

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

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                  
 |                       cos(2*x + 3)
 | sin(2*x + 3) dx = C - ------------
 |                            2      
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∫sin⁡(2x+3) dx=C−cos⁡(2x+3)2\int \sin{\left(2 x + 3 \right)}\, dx = C - \frac{\cos{\left(2 x + 3 \right)}}{2}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
Plot the graph f(x)
The answer [src] 
cos(3)   cos(5)
------ - ------
  2        2
cos⁡(3)2−cos⁡(5)2\frac{\cos{\left(3 \right)}}{2} - \frac{\cos{\left(5 \right)}}{2} 
=
= 
cos(3)   cos(5)
------ - ------
  2        2
cos⁡(3)2−cos⁡(5)2\frac{\cos{\left(3 \right)}}{2} - \frac{\cos{\left(5 \right)}}{2} 
cos(3)/2 - cos(5)/2
Numerical answer [src] 
-0.636827341031836
-0.636827341031836
The graph
Integral of sin(2x+3) dx

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

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