Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of sin(5x)
Integral of 0dx
Integral of -x+4
Integral of (x-2)^2dx
Identical expressions
x*cos(2x+(pi/ four))
x multiply by co sinus of e of (2x plus ( Pi divide by 4))
x multiply by co sinus of e of (2x plus ( Pi divide by four))
xcos(2x+(pi/4))
xcos2x+pi/4
x*cos(2x+(pi divide by 4))
x*cos(2x+(pi/4))dx
Similar expressions
x*cos(2x-(pi/4))

Integral of x*cos(2x+(pi/4)) dx

The solution

You have entered [src]

  1                   
  /                   
 |                    
 |       /      pi\   
 |  x*cos|2*x + --| dx
 |       \      4 /   
 |                    
/                     
0

$$\int\limits_{0}^{1} x \cos{\left(2 x + \frac{\pi}{4} \right)}\, dx$$

Integral(x*cos(2*x + pi/4), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $x \cos{\left(2 x + \frac{\pi}{4} \right)} = x \cos{\left(2 x + \frac{\pi}{4} \right)}$
2. 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 $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x + \frac{\pi}{4} \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 1$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 2 x + \frac{\pi}{4}$ .
    
    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:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
    
    $\frac{\sin{\left(2 x + \frac{\pi}{4} \right)}}{2}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{\sin{\left(2 x + \frac{\pi}{4} \right)}}{2}\, dx = \frac{\int \sin{\left(2 x + \frac{\pi}{4} \right)}\, dx}{2}$
  1. Let $u = 2 x + \frac{\pi}{4}$ .
    
    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:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      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:
    
    $- \frac{\cos{\left(2 x + \frac{\pi}{4} \right)}}{2}$
  So, the result is: $- \frac{\cos{\left(2 x + \frac{\pi}{4} \right)}}{4}$
Method #2
1. 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 $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x + \frac{\pi}{4} \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 1$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 2 x + \frac{\pi}{4}$ .
    
    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:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
    
    $\frac{\sin{\left(2 x + \frac{\pi}{4} \right)}}{2}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{\sin{\left(2 x + \frac{\pi}{4} \right)}}{2}\, dx = \frac{\int \sin{\left(2 x + \frac{\pi}{4} \right)}\, dx}{2}$
  1. Let $u = 2 x + \frac{\pi}{4}$ .
    
    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:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      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:
    
    $- \frac{\cos{\left(2 x + \frac{\pi}{4} \right)}}{2}$
  So, the result is: $- \frac{\cos{\left(2 x + \frac{\pi}{4} \right)}}{4}$
Method #3
1. Rewrite the integrand:
  
  $x \cos{\left(2 x + \frac{\pi}{4} \right)} = x \cos{\left(2 x + \frac{\pi}{4} \right)}$
2. 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 $\operatorname{dv}{\left(x \right)} = \cos{\left(2 x + \frac{\pi}{4} \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 1$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 2 x + \frac{\pi}{4}$ .
    
    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:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      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:
    
    $\frac{\sin{\left(2 x + \frac{\pi}{4} \right)}}{2}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{\sin{\left(2 x + \frac{\pi}{4} \right)}}{2}\, dx = \frac{\int \sin{\left(2 x + \frac{\pi}{4} \right)}\, dx}{2}$
  1. Let $u = 2 x + \frac{\pi}{4}$ .
    
    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:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$
      
      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:
    
    $- \frac{\cos{\left(2 x + \frac{\pi}{4} \right)}}{2}$
  So, the result is: $- \frac{\cos{\left(2 x + \frac{\pi}{4} \right)}}{4}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                            /      pi\        /      pi\
 |                          cos|2*x + --|   x*sin|2*x + --|
 |      /      pi\             \      4 /        \      4 /
 | x*cos|2*x + --| dx = C + ------------- + ---------------
 |      \      4 /                4                2       
 |                                                         
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

   /    pi\              /    pi\
sin|2 + --|     ___   cos|2 + --|
   \    4 /   \/ 2       \    4 /
----------- - ----- + -----------
     2          8          4

$$\frac{\cos{\left(\frac{\pi}{4} + 2 \right)}}{4} - \frac{\sqrt{2}}{8} + \frac{\sin{\left(\frac{\pi}{4} + 2 \right)}}{2}$$

   /    pi\              /    pi\
sin|2 + --|     ___   cos|2 + --|
   \    4 /   \/ 2       \    4 /
----------- - ----- + -----------
     2          8          4

$$\frac{\cos{\left(\frac{\pi}{4} + 2 \right)}}{4} - \frac{\sqrt{2}}{8} + \frac{\sin{\left(\frac{\pi}{4} + 2 \right)}}{2}$$

sin(2 + pi/4)/2 - sqrt(2)/8 + cos(2 + pi/4)/4

Numerical answer [src]

-0.236729288709518

-0.236729288709518

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of x*cos(2x+(pi/4)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3