Integral of e^3xsen2xdx dx

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

Integral(E^3*x*sin(2*x)*1, (x, 0, 1))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

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

$\frac{\left(- 2 x \cos{\left(2 x \right)} + \sin{\left(2 x \right)}\right) e^{3}}{4}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                   
 |                                                    
 |  3                       /sin(2*x)   x*cos(2*x)\  3
 | e *x*sin(2*x)*1 dx = C + |-------- - ----------|*e 
 |                          \   4           2     /   
/

\int e^{3} x \sin{\left(2 x \right)} 1\, dx = C + \left(- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4}\right) e^{3}

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The answer [src]

/  cos(2)   sin(2)\  3
|- ------ + ------|*e 
\    2        4   /

\left(- \frac{\cos{\left(2 \right)}}{2} + \frac{\sin{\left(2 \right)}}{4}\right) e^{3}

=

/  cos(2)   sin(2)\  3
|- ------ + ------|*e 
\    2        4   /

\left(- \frac{\cos{\left(2 \right)}}{2} + \frac{\sin{\left(2 \right)}}{4}\right) e^{3}

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Numerical answer [src]

8.74519808563438

8.74519808563438

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Integral of e^3xsen2xdx dx

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The solution

Method #1

Method #1

Method #2

Method #2