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Identical expressions
(3x+ two)sin4xdx
(3x plus 2) sinus of 4xdx
(3x plus two) sinus of 4xdx
3x+2sin4xdx
Similar expressions
(3x-2)sin4xdx

Integral of (3x+2)sin4xdx dx

The solution

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 |  (3*x + 2)*sin(4*x) dx
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0

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

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

Detail solution

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

$- \frac{3 x \cos{\left(4 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{2}+ \mathrm{constant}$

The answer is:

$- \frac{3 x \cos{\left(4 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$$\int \left(3 x + 2\right) \sin{\left(4 x \right)}\, dx = C - \frac{3 x \cos{\left(4 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{16} - \frac{\cos{\left(4 x \right)}}{2}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1   5*cos(4)   3*sin(4)
- - -------- + --------
2      4          16

$$\frac{3 \sin{\left(4 \right)}}{16} + \frac{1}{2} - \frac{5 \cos{\left(4 \right)}}{4}$$

1   5*cos(4)   3*sin(4)
- - -------- + --------
2      4          16

$$\frac{3 \sin{\left(4 \right)}}{16} + \frac{1}{2} - \frac{5 \cos{\left(4 \right)}}{4}$$

1/2 - 5*cos(4)/4 + 3*sin(4)/16

Numerical answer [src]

1.17515405820928

1.17515405820928

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

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Identical expressions

Similar expressions

Integral of (3x+2)sin4xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3