Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of sin²xcos²x
Integral of ex^2
Integral of (1+2x^2)/(x^2(1+x^2))
Integral of (1+x^4)^(1/2)
Identical expressions
(x+ two)sin5x
(x plus 2) sinus of 5x
(x plus two) sinus of 5x
x+2sin5x
(x+2)sin5xdx
Similar expressions
(x-2)sin5x

Integral of (x+2)sin5x dx

The solution

You have entered [src]

  1                    
  /                    
 |                     
 |  (x + 2)*sin(5*x) dx
 |                     
/                      
0

$$\int\limits_{0}^{1} \left(x + 2\right) \sin{\left(5 x \right)}\, dx$$

Integral((x + 2)*sin(5*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(x + 2\right) \sin{\left(5 x \right)} = x \sin{\left(5 x \right)} + 2 \sin{\left(5 x \right)}$
2. Integrate term-by-term:
  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 $\operatorname{dv}{\left(x \right)} = \sin{\left(5 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 1$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 5 x$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\sin{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(5 x \right)}}{5}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(5 x \right)}}{5}\right)\, dx = - \frac{\int \cos{\left(5 x \right)}\, dx}{5}$
    1. Let $u = 5 x$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\cos{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(5 x \right)}}{5}$
    So, the result is: $- \frac{\sin{\left(5 x \right)}}{25}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \sin{\left(5 x \right)}\, dx = 2 \int \sin{\left(5 x \right)}\, dx$
    1. Let $u = 5 x$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\sin{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(5 x \right)}}{5}$
    So, the result is: $- \frac{2 \cos{\left(5 x \right)}}{5}$
  The result is: $- \frac{x \cos{\left(5 x \right)}}{5} + \frac{\sin{\left(5 x \right)}}{25} - \frac{2 \cos{\left(5 x \right)}}{5}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = x + 2$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(5 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 1$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 5 x$ .
    
    Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
    
    $\int \frac{\sin{\left(u \right)}}{5}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(5 x \right)}}{5}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{\cos{\left(5 x \right)}}{5}\right)\, dx = - \frac{\int \cos{\left(5 x \right)}\, dx}{5}$
  1. Let $u = 5 x$ .
    
    Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
    
    $\int \frac{\cos{\left(u \right)}}{5}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{5}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(5 x \right)}}{5}$
  So, the result is: $- \frac{\sin{\left(5 x \right)}}{25}$
Method #3
1. Rewrite the integrand:
  
  $\left(x + 2\right) \sin{\left(5 x \right)} = x \sin{\left(5 x \right)} + 2 \sin{\left(5 x \right)}$
2. Integrate term-by-term:
  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 $\operatorname{dv}{\left(x \right)} = \sin{\left(5 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 1$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 5 x$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\sin{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(5 x \right)}}{5}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(5 x \right)}}{5}\right)\, dx = - \frac{\int \cos{\left(5 x \right)}\, dx}{5}$
    1. Let $u = 5 x$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\cos{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{5}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(5 x \right)}}{5}$
    So, the result is: $- \frac{\sin{\left(5 x \right)}}{25}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 \sin{\left(5 x \right)}\, dx = 2 \int \sin{\left(5 x \right)}\, dx$
    1. Let $u = 5 x$ .
      
      Then let $du = 5 dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{\sin{\left(u \right)}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(5 x \right)}}{5}$
    So, the result is: $- \frac{2 \cos{\left(5 x \right)}}{5}$
  The result is: $- \frac{x \cos{\left(5 x \right)}}{5} + \frac{\sin{\left(5 x \right)}}{25} - \frac{2 \cos{\left(5 x \right)}}{5}$
Add the constant of integration:

$- \frac{x \cos{\left(5 x \right)}}{5} + \frac{\sin{\left(5 x \right)}}{25} - \frac{2 \cos{\left(5 x \right)}}{5}+ \mathrm{constant}$

The answer is:

$- \frac{x \cos{\left(5 x \right)}}{5} + \frac{\sin{\left(5 x \right)}}{25} - \frac{2 \cos{\left(5 x \right)}}{5}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                            
 |                           2*cos(5*x)   sin(5*x)   x*cos(5*x)
 | (x + 2)*sin(5*x) dx = C - ---------- + -------- - ----------
 |                               5           25          5     
/

$$\int \left(x + 2\right) \sin{\left(5 x \right)}\, dx = C - \frac{x \cos{\left(5 x \right)}}{5} + \frac{\sin{\left(5 x \right)}}{25} - \frac{2 \cos{\left(5 x \right)}}{5}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

2   3*cos(5)   sin(5)
- - -------- + ------
5      5         25

$$- \frac{3 \cos{\left(5 \right)}}{5} + \frac{\sin{\left(5 \right)}}{25} + \frac{2}{5}$$

2   3*cos(5)   sin(5)
- - -------- + ------
5      5         25

$$- \frac{3 \cos{\left(5 \right)}}{5} + \frac{\sin{\left(5 \right)}}{25} + \frac{2}{5}$$

2/5 - 3*cos(5)/5 + sin(5)/25

Numerical answer [src]

0.191445717735539

0.191445717735539

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x+2)sin5x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3