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(7x+ four)*sin*(3x/ eight)
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(7x+4)*sin*(3x/8)dx
Similar expressions
(7x-4)*sin*(3x/8)

Integral of (7x+4)sin(3x/8) dx

The solution

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  1                      
  /                      
 |                       
 |               /3*x\   
 |  (7*x + 4)*sin|---| dx
 |               \ 8 /   
 |                       
/                        
0

$$\int\limits_{0}^{1} \left(7 x + 4\right) \sin{\left(\frac{3 x}{8} \right)}\, dx$$

Integral((7*x + 4)*sin((3*x)/8), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(7 x + 4\right) \sin{\left(\frac{3 x}{8} \right)} = 7 x \sin{\left(\frac{3 x}{8} \right)} + 4 \sin{\left(\frac{3 x}{8} \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 7 x \sin{\left(\frac{3 x}{8} \right)}\, dx = 7 \int x \sin{\left(\frac{3 x}{8} \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{3 x}{8} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{3 x}{8}$ .
      
      Then let $du = \frac{3 dx}{8}$ and substitute $\frac{8 du}{3}$ :
      
      $\int \frac{8 \sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{8 \int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{8 \cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{8 \cos{\left(\frac{3 x}{8} \right)}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{8 \cos{\left(\frac{3 x}{8} \right)}}{3}\right)\, dx = - \frac{8 \int \cos{\left(\frac{3 x}{8} \right)}\, dx}{3}$
      
      Let $u = \frac{3 x}{8}$ .
      
      Then let $du = \frac{3 dx}{8}$ and substitute $\frac{8 du}{3}$ :
      
      $\int \frac{8 \cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{8 \int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{8 \sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{8 \sin{\left(\frac{3 x}{8} \right)}}{3}$
      
      So, the result is: $- \frac{64 \sin{\left(\frac{3 x}{8} \right)}}{9}$
    So, the result is: $- \frac{56 x \cos{\left(\frac{3 x}{8} \right)}}{3} + \frac{448 \sin{\left(\frac{3 x}{8} \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 4 \sin{\left(\frac{3 x}{8} \right)}\, dx = 4 \int \sin{\left(\frac{3 x}{8} \right)}\, dx$
    1. Let $u = \frac{3 x}{8}$ .
      
      Then let $du = \frac{3 dx}{8}$ and substitute $\frac{8 du}{3}$ :
      
      $\int \frac{8 \sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{8 \int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{8 \cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{8 \cos{\left(\frac{3 x}{8} \right)}}{3}$
    So, the result is: $- \frac{32 \cos{\left(\frac{3 x}{8} \right)}}{3}$
  The result is: $- \frac{56 x \cos{\left(\frac{3 x}{8} \right)}}{3} + \frac{448 \sin{\left(\frac{3 x}{8} \right)}}{9} - \frac{32 \cos{\left(\frac{3 x}{8} \right)}}{3}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = 7 x + 4$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{3 x}{8} \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 7$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = \frac{3 x}{8}$ .
    
    Then let $du = \frac{3 dx}{8}$ and substitute $\frac{8 du}{3}$ :
    
    $\int \frac{8 \sin{\left(u \right)}}{3}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{8 \int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{8 \cos{\left(u \right)}}{3}$
    Now substitute $u$ back in:
    
    $- \frac{8 \cos{\left(\frac{3 x}{8} \right)}}{3}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{56 \cos{\left(\frac{3 x}{8} \right)}}{3}\right)\, dx = - \frac{56 \int \cos{\left(\frac{3 x}{8} \right)}\, dx}{3}$
  1. Let $u = \frac{3 x}{8}$ .
    
    Then let $du = \frac{3 dx}{8}$ and substitute $\frac{8 du}{3}$ :
    
    $\int \frac{8 \cos{\left(u \right)}}{3}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{8 \int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{8 \sin{\left(u \right)}}{3}$
    Now substitute $u$ back in:
    
    $\frac{8 \sin{\left(\frac{3 x}{8} \right)}}{3}$
  So, the result is: $- \frac{448 \sin{\left(\frac{3 x}{8} \right)}}{9}$
Method #3
1. Rewrite the integrand:
  
  $\left(7 x + 4\right) \sin{\left(\frac{3 x}{8} \right)} = 7 x \sin{\left(\frac{3 x}{8} \right)} + 4 \sin{\left(\frac{3 x}{8} \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 7 x \sin{\left(\frac{3 x}{8} \right)}\, dx = 7 \int x \sin{\left(\frac{3 x}{8} \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(\frac{3 x}{8} \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = \frac{3 x}{8}$ .
      
      Then let $du = \frac{3 dx}{8}$ and substitute $\frac{8 du}{3}$ :
      
      $\int \frac{8 \sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{8 \int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{8 \cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{8 \cos{\left(\frac{3 x}{8} \right)}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{8 \cos{\left(\frac{3 x}{8} \right)}}{3}\right)\, dx = - \frac{8 \int \cos{\left(\frac{3 x}{8} \right)}\, dx}{3}$
      
      Let $u = \frac{3 x}{8}$ .
      
      Then let $du = \frac{3 dx}{8}$ and substitute $\frac{8 du}{3}$ :
      
      $\int \frac{8 \cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{8 \int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{8 \sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{8 \sin{\left(\frac{3 x}{8} \right)}}{3}$
      
      So, the result is: $- \frac{64 \sin{\left(\frac{3 x}{8} \right)}}{9}$
    So, the result is: $- \frac{56 x \cos{\left(\frac{3 x}{8} \right)}}{3} + \frac{448 \sin{\left(\frac{3 x}{8} \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 4 \sin{\left(\frac{3 x}{8} \right)}\, dx = 4 \int \sin{\left(\frac{3 x}{8} \right)}\, dx$
    1. Let $u = \frac{3 x}{8}$ .
      
      Then let $du = \frac{3 dx}{8}$ and substitute $\frac{8 du}{3}$ :
      
      $\int \frac{8 \sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{8 \int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{8 \cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{8 \cos{\left(\frac{3 x}{8} \right)}}{3}$
    So, the result is: $- \frac{32 \cos{\left(\frac{3 x}{8} \right)}}{3}$
  The result is: $- \frac{56 x \cos{\left(\frac{3 x}{8} \right)}}{3} + \frac{448 \sin{\left(\frac{3 x}{8} \right)}}{9} - \frac{32 \cos{\left(\frac{3 x}{8} \right)}}{3}$
Add the constant of integration:

$- \frac{56 x \cos{\left(\frac{3 x}{8} \right)}}{3} + \frac{448 \sin{\left(\frac{3 x}{8} \right)}}{9} - \frac{32 \cos{\left(\frac{3 x}{8} \right)}}{3}+ \mathrm{constant}$

The answer is:

$- \frac{56 x \cos{\left(\frac{3 x}{8} \right)}}{3} + \frac{448 \sin{\left(\frac{3 x}{8} \right)}}{9} - \frac{32 \cos{\left(\frac{3 x}{8} \right)}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                  /3*x\          /3*x\           /3*x\
 |                             32*cos|---|   448*sin|---|   56*x*cos|---|
 |              /3*x\                \ 8 /          \ 8 /           \ 8 /
 | (7*x + 4)*sin|---| dx = C - ----------- + ------------ - -------------
 |              \ 8 /               3             9               3      
 |                                                                       
/

$$\int \left(7 x + 4\right) \sin{\left(\frac{3 x}{8} \right)}\, dx = C - \frac{56 x \cos{\left(\frac{3 x}{8} \right)}}{3} + \frac{448 \sin{\left(\frac{3 x}{8} \right)}}{9} - \frac{32 \cos{\left(\frac{3 x}{8} \right)}}{3}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

32   88*cos(3/8)   448*sin(3/8)
-- - ----------- + ------------
3         3             9

$$- \frac{88 \cos{\left(\frac{3}{8} \right)}}{3} + \frac{32}{3} + \frac{448 \sin{\left(\frac{3}{8} \right)}}{9}$$

32   88*cos(3/8)   448*sin(3/8)
-- - ----------- + ------------
3         3             9

$$- \frac{88 \cos{\left(\frac{3}{8} \right)}}{3} + \frac{32}{3} + \frac{448 \sin{\left(\frac{3}{8} \right)}}{9}$$

32/3 - 88*cos(3/8)/3 + 448*sin(3/8)/9

Numerical answer [src]

1.60400898285537

1.60400898285537

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (7x+4)sin(3x/8) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (7x+4)*sin*(3x/8) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Integral of (7x+4)sin(3x/8) dx