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sin(4x+1)
Solving integrals/ sin(4x+1)

Integral of sin(4x+1) dx

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The solution

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∫01sin⁡(4x+1) dx\int\limits_{0}^{1} \sin{\left(4 x + 1 \right)}\, dx 
Integral(sin(4*x + 1), (x, 0, 1))
Detail solution

  1. Let u=4x+1u = 4 x + 1.

    Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

    ∫sin⁡(u)4 du\int \frac{\sin{\left(u \right)}}{4}\, du

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

      ∫sin⁡(u) du=∫sin⁡(u) du4\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}

      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)4- \frac{\cos{\left(u \right)}}{4}

    Now substitute uu back in:

    −cos⁡(4x+1)4- \frac{\cos{\left(4 x + 1 \right)}}{4}

  2. Now simplify:

    −cos⁡(4x+1)4- \frac{\cos{\left(4 x + 1 \right)}}{4}

  3. Add the constant of integration:

    −cos⁡(4x+1)4+constant- \frac{\cos{\left(4 x + 1 \right)}}{4}+ \mathrm{constant}

The answer is:

−cos⁡(4x+1)4+constant- \frac{\cos{\left(4 x + 1 \right)}}{4}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                  
 |                       cos(4*x + 1)
 | sin(4*x + 1) dx = C - ------------
 |                            4      
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∫sin⁡(4x+1) dx=C−cos⁡(4x+1)4\int \sin{\left(4 x + 1 \right)}\, dx = C - \frac{\cos{\left(4 x + 1 \right)}}{4}
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(5)   cos(1)
- ------ + ------
    4        4
−cos⁡(5)4+cos⁡(1)4- \frac{\cos{\left(5 \right)}}{4} + \frac{\cos{\left(1 \right)}}{4} 
=
= 
  cos(5)   cos(1)
- ------ + ------
    4        4
−cos⁡(5)4+cos⁡(1)4- \frac{\cos{\left(5 \right)}}{4} + \frac{\cos{\left(1 \right)}}{4} 
-cos(5)/4 + cos(1)/4
Numerical answer [src] 
0.0641600301012284
0.0641600301012284
The graph
Integral of sin(4x+1) dx

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

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