Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of x^4/(x^4-2*x^2+1)
Integral of x²+2x
Integral of dx/1+4x^2
Integral of (ctg(x))^3/sin(2x)
Identical expressions
(4x- three)*sin(x/ two)
(4x minus 3) multiply by sinus of (x divide by 2)
(4x minus three) multiply by sinus of (x divide by two)
(4x-3)sin(x/2)
4x-3sinx/2
(4x-3)*sin(x divide by 2)
(4x-3)*sin(x/2)dx
Similar expressions
(4x+3)*sin(x/2)

Solving integrals/ (4x-3)*sin(x/2)

Integral of (4x-3)*sin(x/2) dx

The solution

You have entered [src]

  1                    
  /                    
 |                     
 |               /x\   
 |  (4*x - 3)*sin|-| dx
 |               \2/   
 |                     
/                      
0

$$\int\limits_{0}^{1} \left(4 x - 3\right) \sin{\left(\frac{x}{2} \right)}\, dx$$

Integral((4*x - 1*3)*sin(x/2), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(4 x - 3\right) \sin{\left(\frac{x}{2} \right)} = 4 x \sin{\left(\frac{x}{2} \right)} - 3 \sin{\left(\frac{x}{2} \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 4 x \sin{\left(\frac{x}{2} \right)}\, dx = 4 \int x \sin{\left(\frac{x}{2} \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{2} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \sin{\left(u \right)}\, du = 2 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 2 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 2 \cos{\left(\frac{x}{2} \right)}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \cos{\left(\frac{x}{2} \right)}\right)\, dx = - 2 \int \cos{\left(\frac{x}{2} \right)}\, dx$
      
      Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \cos{\left(u \right)}\, du = 2 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $2 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $2 \sin{\left(\frac{x}{2} \right)}$
      
      So, the result is: $- 4 \sin{\left(\frac{x}{2} \right)}$
    So, the result is: $- 8 x \cos{\left(\frac{x}{2} \right)} + 16 \sin{\left(\frac{x}{2} \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 3 \sin{\left(\frac{x}{2} \right)}\right)\, dx = - 3 \int \sin{\left(\frac{x}{2} \right)}\, dx$
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \sin{\left(u \right)}\, du = 2 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 2 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 2 \cos{\left(\frac{x}{2} \right)}$
    So, the result is: $6 \cos{\left(\frac{x}{2} \right)}$
  The result is: $- 8 x \cos{\left(\frac{x}{2} \right)} + 16 \sin{\left(\frac{x}{2} \right)} + 6 \cos{\left(\frac{x}{2} \right)}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = 4 x - 3$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{2} \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 4$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = \frac{x}{2}$ .
    
    Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
    
    $\int 4 \sin{\left(u \right)}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \sin{\left(u \right)}\, du = 2 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 2 \cos{\left(u \right)}$
    Now substitute $u$ back in:
    
    $- 2 \cos{\left(\frac{x}{2} \right)}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 8 \cos{\left(\frac{x}{2} \right)}\right)\, dx = - 8 \int \cos{\left(\frac{x}{2} \right)}\, dx$
  1. Let $u = \frac{x}{2}$ .
    
    Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
    
    $\int 4 \cos{\left(u \right)}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \cos{\left(u \right)}\, du = 2 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $2 \sin{\left(u \right)}$
    Now substitute $u$ back in:
    
    $2 \sin{\left(\frac{x}{2} \right)}$
  So, the result is: $- 16 \sin{\left(\frac{x}{2} \right)}$
Method #3
1. Rewrite the integrand:
  
  $\left(4 x - 3\right) \sin{\left(\frac{x}{2} \right)} = 4 x \sin{\left(\frac{x}{2} \right)} - 3 \sin{\left(\frac{x}{2} \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 4 x \sin{\left(\frac{x}{2} \right)}\, dx = 4 \int x \sin{\left(\frac{x}{2} \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{x}{2} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \sin{\left(u \right)}\, du = 2 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 2 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 2 \cos{\left(\frac{x}{2} \right)}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \cos{\left(\frac{x}{2} \right)}\right)\, dx = - 2 \int \cos{\left(\frac{x}{2} \right)}\, dx$
      
      Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 \cos{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \cos{\left(u \right)}\, du = 2 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $2 \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $2 \sin{\left(\frac{x}{2} \right)}$
      
      So, the result is: $- 4 \sin{\left(\frac{x}{2} \right)}$
    So, the result is: $- 8 x \cos{\left(\frac{x}{2} \right)} + 16 \sin{\left(\frac{x}{2} \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 3 \sin{\left(\frac{x}{2} \right)}\right)\, dx = - 3 \int \sin{\left(\frac{x}{2} \right)}\, dx$
    1. Let $u = \frac{x}{2}$ .
      
      Then let $du = \frac{dx}{2}$ and substitute $2 du$ :
      
      $\int 4 \sin{\left(u \right)}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 \sin{\left(u \right)}\, du = 2 \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- 2 \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- 2 \cos{\left(\frac{x}{2} \right)}$
    So, the result is: $6 \cos{\left(\frac{x}{2} \right)}$
  The result is: $- 8 x \cos{\left(\frac{x}{2} \right)} + 16 \sin{\left(\frac{x}{2} \right)} + 6 \cos{\left(\frac{x}{2} \right)}$
Add the constant of integration:

$- 8 x \cos{\left(\frac{x}{2} \right)} + 16 \sin{\left(\frac{x}{2} \right)} + 6 \cos{\left(\frac{x}{2} \right)}+ \mathrm{constant}$

The answer is:

$- 8 x \cos{\left(\frac{x}{2} \right)} + 16 \sin{\left(\frac{x}{2} \right)} + 6 \cos{\left(\frac{x}{2} \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                           
 |                                                            
 |              /x\               /x\         /x\          /x\
 | (4*x - 3)*sin|-| dx = C + 6*cos|-| + 16*sin|-| - 8*x*cos|-|
 |              \2/               \2/         \2/          \2/
 |                                                            
/

$$2\,\left(8\,\left(\sin \left({{x}\over{2}}\right)-{{\cos \left({{x }\over{2}}\right)\,x}\over{2}}\right)+3\,\cos \left({{x}\over{2}} \right)\right)$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-6 - 2*cos(1/2) + 16*sin(1/2)

$$16\,\sin \left({{1}\over{2}}\right)-2\,\cos \left({{1}\over{2}} \right)-6$$

-6 - 2*cos(1/2) + 16*sin(1/2)

$$-6 - 2 \cos{\left(\frac{1}{2} \right)} + 16 \sin{\left(\frac{1}{2} \right)}$$

Simplify

Numerical answer [src]

-0.0843565061134974

-0.0843565061134974

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (4x-3)*sin(x/2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3