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(4x+ seven)*cos^3x
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(4x+7)*cos3x
4x+7*cos3x
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(4x+7)*cos to the power of 3x
(4x+7)cos^3x
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Similar expressions
(4x-7)*cos^3x

Integral of (4x+7)*cos^3x dx

The solution

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

Integral((4*x + 7)*cos(x)^3, (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(4 x + 7\right) \cos^{3}{\left(x \right)} = 4 x \cos^{3}{\left(x \right)} + 7 \cos^{3}{\left(x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 4 x \cos^{3}{\left(x \right)}\, dx = 4 \int x \cos^{3}{\left(x \right)}\, dx$
    1. Don't know the steps in finding this integral.
      
      But the integral is
      
      $\frac{2 x \sin^{3}{\left(x \right)}}{3} + x \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \frac{2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}}{3} + \frac{7 \cos^{3}{\left(x \right)}}{9}$
    So, the result is: $\frac{8 x \sin^{3}{\left(x \right)}}{3} + 4 x \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \frac{8 \sin^{2}{\left(x \right)} \cos{\left(x \right)}}{3} + \frac{28 \cos^{3}{\left(x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 7 \cos^{3}{\left(x \right)}\, dx = 7 \int \cos^{3}{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}$
    2. Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $- \frac{u^{3}}{3}$
      
      The result is: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
    So, the result is: $- \frac{7 \sin^{3}{\left(x \right)}}{3} + 7 \sin{\left(x \right)}$
  The result is: $\frac{8 x \sin^{3}{\left(x \right)}}{3} + 4 x \sin{\left(x \right)} \cos^{2}{\left(x \right)} - \frac{7 \sin^{3}{\left(x \right)}}{3} + \frac{8 \sin^{2}{\left(x \right)} \cos{\left(x \right)}}{3} + 7 \sin{\left(x \right)} + \frac{28 \cos^{3}{\left(x \right)}}{9}$
Method #2
1. Rewrite the integrand:
  
  $\left(4 x + 7\right) \cos^{3}{\left(x \right)} = 4 x \cos^{3}{\left(x \right)} + 7 \cos^{3}{\left(x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 4 x \cos^{3}{\left(x \right)}\, dx = 4 \int x \cos^{3}{\left(x \right)}\, dx$
    1. Don't know the steps in finding this integral.
      
      But the integral is
      
      $\frac{2 x \sin^{3}{\left(x \right)}}{3} + x \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \frac{2 \sin^{2}{\left(x \right)} \cos{\left(x \right)}}{3} + \frac{7 \cos^{3}{\left(x \right)}}{9}$
    So, the result is: $\frac{8 x \sin^{3}{\left(x \right)}}{3} + 4 x \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \frac{8 \sin^{2}{\left(x \right)} \cos{\left(x \right)}}{3} + \frac{28 \cos^{3}{\left(x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 7 \cos^{3}{\left(x \right)}\, dx = 7 \int \cos^{3}{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}$
    2. Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $- \frac{u^{3}}{3}$
      
      The result is: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$
    So, the result is: $- \frac{7 \sin^{3}{\left(x \right)}}{3} + 7 \sin{\left(x \right)}$
  The result is: $\frac{8 x \sin^{3}{\left(x \right)}}{3} + 4 x \sin{\left(x \right)} \cos^{2}{\left(x \right)} - \frac{7 \sin^{3}{\left(x \right)}}{3} + \frac{8 \sin^{2}{\left(x \right)} \cos{\left(x \right)}}{3} + 7 \sin{\left(x \right)} + \frac{28 \cos^{3}{\left(x \right)}}{9}$
Now simplify:

$- \frac{4 x \sin^{3}{\left(x \right)}}{3} + 4 x \sin{\left(x \right)} - \frac{7 \sin^{3}{\left(x \right)}}{3} + 7 \sin{\left(x \right)} + \frac{4 \cos^{3}{\left(x \right)}}{9} + \frac{8 \cos{\left(x \right)}}{3}$
Add the constant of integration:

$- \frac{4 x \sin^{3}{\left(x \right)}}{3} + 4 x \sin{\left(x \right)} - \frac{7 \sin^{3}{\left(x \right)}}{3} + 7 \sin{\left(x \right)} + \frac{4 \cos^{3}{\left(x \right)}}{9} + \frac{8 \cos{\left(x \right)}}{3}+ \mathrm{constant}$

The answer is:

$- \frac{4 x \sin^{3}{\left(x \right)}}{3} + 4 x \sin{\left(x \right)} - \frac{7 \sin^{3}{\left(x \right)}}{3} + 7 \sin{\left(x \right)} + \frac{4 \cos^{3}{\left(x \right)}}{9} + \frac{8 \cos{\left(x \right)}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                                                  
 |                                            3            3             3           2                               
 |              3                        7*sin (x)   28*cos (x)   8*x*sin (x)   8*sin (x)*cos(x)          2          
 | (4*x + 7)*cos (x) dx = C + 7*sin(x) - --------- + ---------- + ----------- + ---------------- + 4*x*cos (x)*sin(x)
 |                                           3           9             3               3                             
/

$$\int \left(4 x + 7\right) \cos^{3}{\left(x \right)}\, dx = C + \frac{8 x \sin^{3}{\left(x \right)}}{3} + 4 x \sin{\left(x \right)} \cos^{2}{\left(x \right)} - \frac{7 \sin^{3}{\left(x \right)}}{3} + \frac{8 \sin^{2}{\left(x \right)} \cos{\left(x \right)}}{3} + 7 \sin{\left(x \right)} + \frac{28 \cos^{3}{\left(x \right)}}{9}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

             3            3                               2          
  28   22*sin (1)   28*cos (1)         2             8*sin (1)*cos(1)
- -- + ---------- + ---------- + 11*cos (1)*sin(1) + ----------------
  9        3            9                                   3

$$- \frac{28}{9} + \frac{28 \cos^{3}{\left(1 \right)}}{9} + \frac{8 \sin^{2}{\left(1 \right)} \cos{\left(1 \right)}}{3} + 11 \sin{\left(1 \right)} \cos^{2}{\left(1 \right)} + \frac{22 \sin^{3}{\left(1 \right)}}{3}$$

             3            3                               2          
  28   22*sin (1)   28*cos (1)         2             8*sin (1)*cos(1)
- -- + ---------- + ---------- + 11*cos (1)*sin(1) + ----------------
  9        3            9                                   3

-28/9 + 22*sin(1)^3/3 + 28*cos(1)^3/9 + 11*cos(1)^2*sin(1) + 8*sin(1)^2*cos(1)/3

Numerical answer [src]

5.47129227225773

5.47129227225773

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

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How to use it?

Identical expressions

Similar expressions

Integral of (4x+7)*cos^3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2