Integral of sin(2x+1) dx

The solution

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  1                
  /                
 |                 
 |  sin(2*x + 1) dx
 |                 
/                  
0

$$\int\limits_{0}^{1} \sin{\left(2 x + 1 \right)}\, dx$$

Integral(sin(2*x + 1), (x, 0, 1))

Detail solution

Let $u = 2 x + 1$ .

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

$\int \frac{\sin{\left(u \right)}}{2}\, du$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
  1. The integral of sine is negative cosine:
    
    $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
  So, the result is: $- \frac{\cos{\left(u \right)}}{2}$
Now substitute $u$ back in:

$- \frac{\cos{\left(2 x + 1 \right)}}{2}$
Now simplify:

$- \frac{\cos{\left(2 x + 1 \right)}}{2}$
Add the constant of integration:

$- \frac{\cos{\left(2 x + 1 \right)}}{2}+ \mathrm{constant}$

The answer is:

$- \frac{\cos{\left(2 x + 1 \right)}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                  
 |                       cos(2*x + 1)
 | sin(2*x + 1) dx = C - ------------
 |                            2      
/

$$\int \sin{\left(2 x + 1 \right)}\, dx = C - \frac{\cos{\left(2 x + 1 \right)}}{2}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

cos(1)   cos(3)
------ - ------
  2        2

$$\frac{\cos{\left(1 \right)}}{2} - \frac{\cos{\left(3 \right)}}{2}$$

cos(1)   cos(3)
------ - ------
  2        2

$$\frac{\cos{\left(1 \right)}}{2} - \frac{\cos{\left(3 \right)}}{2}$$

cos(1)/2 - cos(3)/2

Numerical answer [src]

0.765147401234293

0.765147401234293

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(2x+1) dx

Limits of integration:

The graph:

Piecewise:

The solution