Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of x^(-2)
Integral of sin(3x)
Integral of logx
Integral of x^2/(1+x)
Identical expressions
xcos(2x+ one)
x co sinus of e of (2x plus 1)
x co sinus of e of (2x plus one)
xcos2x+1
xcos(2x+1)dx
Similar expressions
xcos(2x-1)
(1-6x)cos(2x+1)

Integral of xcos(2x+1) dx

The solution

You have entered [src]

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

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

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

Detail solution

Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x + 1 \right)}$ .

Then $\operatorname{du}{\left(x \right)} = 1$ .

To find $v{\left(x \right)}$ :
1. 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 evaluate the sub-integral.
The integral of a constant times a function is the constant times the integral of the function:

$\int \frac{\sin{\left(2 x + 1 \right)}}{2}\, dx = \frac{\int \sin{\left(2 x + 1 \right)}\, dx}{2}$
1. 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:
  
  $- \frac{\cos{\left(2 x + 1 \right)}}{2}$
So, the result is: $- \frac{\cos{\left(2 x + 1 \right)}}{4}$
Add the constant of integration:

$\frac{x \sin{\left(2 x + 1 \right)}}{2} + \frac{\cos{\left(2 x + 1 \right)}}{4}+ \mathrm{constant}$

The answer is:

$\frac{x \sin{\left(2 x + 1 \right)}}{2} + \frac{\cos{\left(2 x + 1 \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                     
 |                         cos(1 + 2*x)   x*sin(1 + 2*x)
 | x*cos(2*x + 1) dx = C + ------------ + --------------
 |                              4               2       
/

$$\int x \cos{\left(2 x + 1 \right)}\, dx = C + \frac{x \sin{\left(2 x + 1 \right)}}{2} + \frac{\cos{\left(2 x + 1 \right)}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$0$$

Numerical answer [src]

0.0

0.0

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of xcos(2x+1) dx

Limits of integration:

The graph:

Piecewise:

The solution