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Integral of d{x}:
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Integral of (x)/(1+x^2)
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Integral of dx/(9*x^2-1)
Identical expressions
three ^cos4x*sin4x
3 to the power of co sinus of e of 4x multiply by sinus of 4x
three to the power of co sinus of e of 4x multiply by sinus of 4x
3cos4x*sin4x
3^cos4xsin4x
3cos4xsin4x
3^cos4x*sin4xdx

Integral of 3^cos4x*sin4x dx

The solution

You have entered [src]

  1                      
  /                      
 |                       
 |   cos(4*x)            
 |  3        *sin(4*x) dx
 |                       
/                        
0

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

Integral(3^cos(4*x)*sin(4*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \cos{\left(4 x \right)}$ .
  
  Then let $du = - 4 \sin{\left(4 x \right)} dx$ and substitute $- \frac{du}{4}$ :
  
  $\int \left(- \frac{3^{u}}{4}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3^{u}\, du = - \frac{\int 3^{u}\, du}{4}$
    1. The integral of an exponential function is itself divided by the natural logarithm of the base.
      
      $\int 3^{u}\, du = \frac{3^{u}}{\log{\left(3 \right)}}$
    So, the result is: $- \frac{3^{u}}{4 \log{\left(3 \right)}}$
  Now substitute $u$ back in:
  
  $- \frac{3^{\cos{\left(4 x \right)}}}{4 \log{\left(3 \right)}}$
Method #2
1. Let $u = 4 x$ .
  
  Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
  
  $\int \frac{3^{\cos{\left(u \right)}} \sin{\left(u \right)}}{4}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 3^{\cos{\left(u \right)}} \sin{\left(u \right)}\, du = \frac{\int 3^{\cos{\left(u \right)}} \sin{\left(u \right)}\, du}{4}$
    1. Let $u = \cos{\left(u \right)}$ .
      
      Then let $du = - \sin{\left(u \right)} du$ and substitute $- du$ :
      
      $\int \left(- 3^{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3^{u}\, du = - \int 3^{u}\, du$
      
      The integral of an exponential function is itself divided by the natural logarithm of the base.
      
      $\int 3^{u}\, du = \frac{3^{u}}{\log{\left(3 \right)}}$
      
      So, the result is: $- \frac{3^{u}}{\log{\left(3 \right)}}$
      
      Now substitute $u$ back in:
      
      $- \frac{3^{\cos{\left(u \right)}}}{\log{\left(3 \right)}}$
    So, the result is: $- \frac{3^{\cos{\left(u \right)}}}{4 \log{\left(3 \right)}}$
  Now substitute $u$ back in:
  
  $- \frac{3^{\cos{\left(4 x \right)}}}{4 \log{\left(3 \right)}}$
Add the constant of integration:

$- \frac{3^{\cos{\left(4 x \right)}}}{4 \log{\left(3 \right)}}+ \mathrm{constant}$

The answer is:

$- \frac{3^{\cos{\left(4 x \right)}}}{4 \log{\left(3 \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                     
 |                              cos(4*x)
 |  cos(4*x)                   3        
 | 3        *sin(4*x) dx = C - ---------
 |                              4*log(3)
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

            cos(4) 
   3       3       
-------- - --------
4*log(3)   4*log(3)

$$- \frac{1}{4 \cdot 3^{- \cos{\left(4 \right)}} \log{\left(3 \right)}} + \frac{3}{4 \log{\left(3 \right)}}$$

            cos(4) 
   3       3       
-------- - --------
4*log(3)   4*log(3)

$$- \frac{1}{4 \cdot 3^{- \cos{\left(4 \right)}} \log{\left(3 \right)}} + \frac{3}{4 \log{\left(3 \right)}}$$

3/(4*log(3)) - 3^cos(4)/(4*log(3))

Numerical answer [src]

0.571703618146759

0.571703618146759

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

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Identical expressions

Integral of 3^cos4x*sin4x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2