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e^(3sinx+ one)cosx
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e to the power of (3 sinus of x plus one) co sinus of e of x
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Similar expressions
e^(3sinx-1)cosx

Solving integrals/ e^(3sinx+1)cosx

Integral of e^(3sinx+1)cosx dx

The solution

You have entered [src]

  1                        
  /                        
 |                         
 |   3*sin(x) + 1          
 |  e            *cos(x) dx
 |                         
/                          
0

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

Integral(E^(3*sin(x) + 1)*cos(x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $e^{3 \sin{\left(x \right)} + 1} \cos{\left(x \right)} = e e^{3 \sin{\left(x \right)}} \cos{\left(x \right)}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int e e^{3 \sin{\left(x \right)}} \cos{\left(x \right)}\, dx = e \int e^{3 \sin{\left(x \right)}} \cos{\left(x \right)}\, dx$
  1. Let $u = e^{3 \sin{\left(x \right)}}$ .
    
    Then let $du = 3 e^{3 \sin{\left(x \right)}} \cos{\left(x \right)} dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{1}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{3}$
    Now substitute $u$ back in:
    
    $\frac{e^{3 \sin{\left(x \right)}}}{3}$
  So, the result is: $\frac{e e^{3 \sin{\left(x \right)}}}{3}$
Method #2
1. Rewrite the integrand:
  
  $e^{3 \sin{\left(x \right)} + 1} \cos{\left(x \right)} = e e^{3 \sin{\left(x \right)}} \cos{\left(x \right)}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int e e^{3 \sin{\left(x \right)}} \cos{\left(x \right)}\, dx = e \int e^{3 \sin{\left(x \right)}} \cos{\left(x \right)}\, dx$
  1. Let $u = e^{3 \sin{\left(x \right)}}$ .
    
    Then let $du = 3 e^{3 \sin{\left(x \right)}} \cos{\left(x \right)} dx$ and substitute $\frac{du}{3}$ :
    
    $\int \frac{1}{9}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{3}$
    Now substitute $u$ back in:
    
    $\frac{e^{3 \sin{\left(x \right)}}}{3}$
  So, the result is: $\frac{e e^{3 \sin{\left(x \right)}}}{3}$
Now simplify:

$\frac{e^{3 \sin{\left(x \right)} + 1}}{3}$
Add the constant of integration:

$\frac{e^{3 \sin{\left(x \right)} + 1}}{3}+ \mathrm{constant}$

The answer is:

$\frac{e^{3 \sin{\left(x \right)} + 1}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                         
 |                                  3*sin(x)
 |  3*sin(x) + 1                 e*e        
 | e            *cos(x) dx = C + -----------
 |                                    3     
/

$$\int e^{3 \sin{\left(x \right)} + 1} \cos{\left(x \right)}\, dx = C + \frac{e e^{3 \sin{\left(x \right)}}}{3}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

         3*sin(1)
  e   e*e        
- - + -----------
  3        3

$$- \frac{e}{3} + \frac{e e^{3 \sin{\left(1 \right)}}}{3}$$

         3*sin(1)
  e   e*e        
- - + -----------
  3        3

$$- \frac{e}{3} + \frac{e e^{3 \sin{\left(1 \right)}}}{3}$$

Упростить

Numerical answer [src]

10.4051884082202

10.4051884082202

The graph

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

Other calculators

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Identical expressions

Similar expressions

Integral of e^(3sinx+1)cosx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2