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(sin^ three)xcosxdx
( sinus of cubed )x co sinus of e of xdx
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sin3xcosxdx
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sin^3xcosxdx

Solving integrals/ (sin^3)xcosxdx

Integral of (sin^3)xcosxdx dx

The solution

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 |  sin (x)*x*cos(x)*1 dx
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0

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

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

Detail solution

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 $\operatorname{dv}{\left(x \right)} = \sin^{3}{\left(x \right)} \cos{\left(x \right)}$ .

Then $\operatorname{du}{\left(x \right)} = 1$ .

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

$\int \frac{\sin^{4}{\left(x \right)}}{4}\, dx = \frac{\int \sin^{4}{\left(x \right)}\, dx}{4}$
1. Rewrite the integrand:
  
  $\sin^{4}{\left(x \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2}$
2. There are multiple ways to do this integral.
  Method #1
  1. Rewrite the integrand:
    
    $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{4}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
      
      So, the result is: $\frac{x}{8} + \frac{\sin{\left(4 x \right)}}{32}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      
      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: $- \frac{\sin{\left(2 x \right)}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, dx = \frac{x}{4}$
    The result is: $\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}$
  Method #2
  1. Rewrite the integrand:
    
    $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} = \frac{\cos^{2}{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{4}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{2}{\left(2 x \right)}}{4}\, dx = \frac{\int \cos^{2}{\left(2 x \right)}\, dx}{4}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
      
      So, the result is: $\frac{x}{8} + \frac{\sin{\left(4 x \right)}}{32}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      
      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: $- \frac{\sin{\left(2 x \right)}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, dx = \frac{x}{4}$
    The result is: $\frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}$
So, the result is: $\frac{3 x}{32} - \frac{\sin{\left(2 x \right)}}{16} + \frac{\sin{\left(4 x \right)}}{128}$
Add the constant of integration:

$\frac{x \sin^{4}{\left(x \right)}}{4} - \frac{3 x}{32} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{128}+ \mathrm{constant}$

The answer is:

$\frac{x \sin^{4}{\left(x \right)}}{4} - \frac{3 x}{32} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{128}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                 
 |                                                              4   
 |    3                        3*x   sin(4*x)   sin(2*x)   x*sin (x)
 | sin (x)*x*cos(x)*1 dx = C - --- - -------- + -------- + ---------
 |                              32     128         16          4    
/

$$-{{\sin \left(4\,x\right)-4\,x\,\cos \left(4\,x\right)-8\,\sin \left(2\,x\right)+16\,x\,\cos \left(2\,x\right)}\over{128}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

       4           4           2       2           3                  3          
  3*cos (1)   5*sin (1)   3*cos (1)*sin (1)   3*cos (1)*sin(1)   5*sin (1)*cos(1)
- --------- + --------- - ----------------- + ---------------- + ----------------
      32          32              16                 32                 32

$$-{{\sin 4-4\,\cos 4-8\,\sin 2+16\,\cos 2}\over{128}}$$

       4           4           2       2           3                  3          
  3*cos (1)   5*sin (1)   3*cos (1)*sin (1)   3*cos (1)*sin(1)   5*sin (1)*cos(1)
- --------- + --------- - ----------------- + ---------------- + ----------------
      32          32              16                 32                 32

$$- \frac{3 \sin^{2}{\left(1 \right)} \cos^{2}{\left(1 \right)}}{16} - \frac{3 \cos^{4}{\left(1 \right)}}{32} + \frac{3 \sin{\left(1 \right)} \cos^{3}{\left(1 \right)}}{32} + \frac{5 \sin^{3}{\left(1 \right)} \cos{\left(1 \right)}}{32} + \frac{5 \sin^{4}{\left(1 \right)}}{32}$$

Simplify

Numerical answer [src]

0.0943356000876032

0.0943356000876032

The graph

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

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How to use it?

Identical expressions

Integral of (sin^3)xcosxdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2