Integral of cos(2*x+1) dx

The solution

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  1                
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 |  cos(2*x + 1) dx
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0

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

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

Detail solution

Let $u = 2 x + 1$ .

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

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

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

sin(3)   sin(1)
------ - ------
  2        2

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

sin(3)   sin(1)
------ - ------
  2        2

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

sin(3)/2 - sin(1)/2

Numerical answer [src]

-0.350175488374015

-0.350175488374015

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

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Integral of cos(2*x+1) dx

Limits of integration:

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The solution