Integral of sin(4x+1) dx

The solution

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  1                
  /                
 |                 
 |  sin(4*x + 1) dx
 |                 
/                  
0

$$\int\limits_{0}^{1} \sin{\left(4 x + 1 \right)}\, dx$$

Integral(sin(4*x + 1), (x, 0, 1))

Detail solution

Let $u = 4 x + 1$ .

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

$\int \frac{\sin{\left(u \right)}}{4}\, du$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
  1. The integral of sine is negative cosine:
    
    $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
  So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
Now substitute $u$ back in:

$- \frac{\cos{\left(4 x + 1 \right)}}{4}$
Now simplify:

$- \frac{\cos{\left(4 x + 1 \right)}}{4}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                  
 |                       cos(4*x + 1)
 | sin(4*x + 1) dx = C - ------------
 |                            4      
/

$$\int \sin{\left(4 x + 1 \right)}\, dx = C - \frac{\cos{\left(4 x + 1 \right)}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  cos(5)   cos(1)
- ------ + ------
    4        4

$$- \frac{\cos{\left(5 \right)}}{4} + \frac{\cos{\left(1 \right)}}{4}$$

  cos(5)   cos(1)
- ------ + ------
    4        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

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

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Identical expressions

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Integral of sin(4x+1) dx

Limits of integration:

The graph:

Piecewise:

The solution