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Integral of (1-x^2)^(1/2)
Identical expressions
(x- seven)*(-sin(3x)+5cos(3x))
(x minus 7) multiply by ( minus sinus of (3x) plus 5 co sinus of e of (3x))
(x minus seven) multiply by ( minus sinus of (3x) plus 5 co sinus of e of (3x))
(x-7)(-sin(3x)+5cos(3x))
x-7-sin3x+5cos3x
(x-7)*(-sin(3x)+5cos(3x))dx
Similar expressions
(x-7)*(sin(3x)+5cos(3x))
(x+7)*(-sin(3x)+5cos(3x))
(x-7)*(-sin(3x)-5cos(3x))

Solving integrals/ (x-7)*(-sin(3x)+5cos(3x))

Integral of (x-7)*(-sin(3x)+5cos(3x)) dx

The solution

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18

$$\int\limits_{\frac{\pi}{18}}^{\frac{\pi}{4}} \left(x - 7\right) \left(- \sin{\left(3 x \right)} + 5 \cos{\left(3 x \right)}\right)\, dx$$

Integral((x - 7)*(-sin(3*x) + 5*cos(3*x)), (x, pi/18, pi/4))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(x - 7\right) \left(- \sin{\left(3 x \right)} + 5 \cos{\left(3 x \right)}\right) = - x \sin{\left(3 x \right)} + 5 x \cos{\left(3 x \right)} + 7 \sin{\left(3 x \right)} - 35 \cos{\left(3 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- x \sin{\left(3 x \right)}\right)\, dx = - \int x \sin{\left(3 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(3 x \right)}}{3}\right)\, dx = - \frac{\int \cos{\left(3 x \right)}\, dx}{3}$
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      So, the result is: $- \frac{\sin{\left(3 x \right)}}{9}$
    So, the result is: $\frac{x \cos{\left(3 x \right)}}{3} - \frac{\sin{\left(3 x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 5 x \cos{\left(3 x \right)}\, dx = 5 \int x \cos{\left(3 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(3 x \right)}}{3}\, dx = \frac{\int \sin{\left(3 x \right)}\, dx}{3}$
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      So, the result is: $- \frac{\cos{\left(3 x \right)}}{9}$
    So, the result is: $\frac{5 x \sin{\left(3 x \right)}}{3} + \frac{5 \cos{\left(3 x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 7 \sin{\left(3 x \right)}\, dx = 7 \int \sin{\left(3 x \right)}\, dx$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    So, the result is: $- \frac{7 \cos{\left(3 x \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 35 \cos{\left(3 x \right)}\right)\, dx = - 35 \int \cos{\left(3 x \right)}\, dx$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
    So, the result is: $- \frac{35 \sin{\left(3 x \right)}}{3}$
  The result is: $\frac{5 x \sin{\left(3 x \right)}}{3} + \frac{x \cos{\left(3 x \right)}}{3} - \frac{106 \sin{\left(3 x \right)}}{9} - \frac{16 \cos{\left(3 x \right)}}{9}$
Method #2
1. Rewrite the integrand:
  
  $\left(x - 7\right) \left(- \sin{\left(3 x \right)} + 5 \cos{\left(3 x \right)}\right) = - x \sin{\left(3 x \right)} + 5 x \cos{\left(3 x \right)} + 7 \sin{\left(3 x \right)} - 35 \cos{\left(3 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- x \sin{\left(3 x \right)}\right)\, dx = - \int x \sin{\left(3 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(3 x \right)}}{3}\right)\, dx = - \frac{\int \cos{\left(3 x \right)}\, dx}{3}$
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      So, the result is: $- \frac{\sin{\left(3 x \right)}}{9}$
    So, the result is: $\frac{x \cos{\left(3 x \right)}}{3} - \frac{\sin{\left(3 x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 5 x \cos{\left(3 x \right)}\, dx = 5 \int x \cos{\left(3 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(3 x \right)}}{3}\, dx = \frac{\int \sin{\left(3 x \right)}\, dx}{3}$
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      So, the result is: $- \frac{\cos{\left(3 x \right)}}{9}$
    So, the result is: $\frac{5 x \sin{\left(3 x \right)}}{3} + \frac{5 \cos{\left(3 x \right)}}{9}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 7 \sin{\left(3 x \right)}\, dx = 7 \int \sin{\left(3 x \right)}\, dx$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
    So, the result is: $- \frac{7 \cos{\left(3 x \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 35 \cos{\left(3 x \right)}\right)\, dx = - 35 \int \cos{\left(3 x \right)}\, dx$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
    So, the result is: $- \frac{35 \sin{\left(3 x \right)}}{3}$
  The result is: $\frac{5 x \sin{\left(3 x \right)}}{3} + \frac{x \cos{\left(3 x \right)}}{3} - \frac{106 \sin{\left(3 x \right)}}{9} - \frac{16 \cos{\left(3 x \right)}}{9}$
Add the constant of integration:

$\frac{5 x \sin{\left(3 x \right)}}{3} + \frac{x \cos{\left(3 x \right)}}{3} - \frac{106 \sin{\left(3 x \right)}}{9} - \frac{16 \cos{\left(3 x \right)}}{9}+ \mathrm{constant}$

The answer is:

$\frac{5 x \sin{\left(3 x \right)}}{3} + \frac{x \cos{\left(3 x \right)}}{3} - \frac{106 \sin{\left(3 x \right)}}{9} - \frac{16 \cos{\left(3 x \right)}}{9}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                                
 |                                           106*sin(3*x)   16*cos(3*x)   x*cos(3*x)   5*x*sin(3*x)
 | (x - 7)*(-sin(3*x) + 5*cos(3*x)) dx = C - ------------ - ----------- + ---------- + ------------
 |                                                9              9            3             3      
/

$$\int \left(x - 7\right) \left(- \sin{\left(3 x \right)} + 5 \cos{\left(3 x \right)}\right)\, dx = C + \frac{5 x \sin{\left(3 x \right)}}{3} + \frac{x \cos{\left(3 x \right)}}{3} - \frac{106 \sin{\left(3 x \right)}}{9} - \frac{16 \cos{\left(3 x \right)}}{9}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                          ___        ___        ___
53       ___   5*pi   8*\/ 3    pi*\/ 3    pi*\/ 2 
-- - 5*\/ 2  - ---- + ------- - -------- + --------
9              108       9        108         6

$$- 5 \sqrt{2} - \frac{5 \pi}{108} - \frac{\sqrt{3} \pi}{108} + \frac{\sqrt{2} \pi}{6} + \frac{8 \sqrt{3}}{9} + \frac{53}{9}$$

                          ___        ___        ___
53       ___   5*pi   8*\/ 3    pi*\/ 3    pi*\/ 2 
-- - 5*\/ 2  - ---- + ------- - -------- + --------
9              108       9        108         6

$$- 5 \sqrt{2} - \frac{5 \pi}{108} - \frac{\sqrt{3} \pi}{108} + \frac{\sqrt{2} \pi}{6} + \frac{8 \sqrt{3}}{9} + \frac{53}{9}$$

53/9 - 5*sqrt(2) - 5*pi/108 + 8*sqrt(3)/9 - pi*sqrt(3)/108 + pi*sqrt(2)/6

Numerical answer [src]

0.902074864549442

0.902074864549442

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

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Integral of (x-7)*(-sin(3x)+5cos(3x)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2