Problems about integration

• READ THIS FIRST
  
• Integral calculation
  
  • Level 1
    
  • Level 3
    
• External links to integral calculation
  

READ THIS FIRST

If a problem is solved. It is not 'the' answer.
No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Integral calculation

Level 1

• 
  

  / 2 ___ | x V x | ------- dx | 1/3 / x

  
  

   
  
    /  13/6        6   19:6
  = | x     dx  =  -- x     + C
    /              19
  
  

• 
  

  / = | ln4(x) | ------ dx / 5 x

  
  

   
  say ln(x) = u , then du = (1/x) dx
  
    /      4        5           5
  = |   u          u          ln (x)
    | ----- du  = ---- + C =  ------ + C
    /   5          25           25
  
  

• 
  

  / |sqrt(3x-7)dx /

  
  

   
    1 /                      2
  = - |sqrt(3x-7)d(3x-7) =   - (3 x - 7)3/2 +C
    3 /                      9
  
  
  

• 
  

  / | x2 = | --------- dx | _______ / | 6 \| 9 - x

  
  

   
  say x3 = t then 3x2dx = dt
  
                                                    3
    / 1    dt         1         t       1          x
  = | - ---------  =  -  arcsin(-)   =  -  arcsin(---) + C
    |      _______    3         3       3          3
    |     |      2
    / 3  \| 9 - t
  
  

• 
  

  / | |x2.tan(x3+ 1) dx | /

  
  

   
  say x3+ 1 = u  then du = 3 x2dx
  
      /                /
   1  |             1  |  sin(u)
  =-  | tan(u) du = -  | -------du
   3  |             3  |  cos(u)
      /                /
  
  say cos(u) = t  then dt = -sin(u)du
  
        /
     1  |  dt        1                 1
  =- -  |  ---  =  - -  ln|t| +C =   - -  ln|cos(u)| +C
     3  |   t        3                 3
        /
  
         1
     = - -  ln|cos(x3 + 1)| +C
         3
  

• 
  

  / | 2x + arctan(x) |----------------dx | 1 + x2 /

  
  

   
      /                    /
      |    2  x            |  arctan(x)
  =   |-------------dx +   |-------------dx
      |  1  + x.x          | 1  + x2
      /                    /
  
  
  In the first part say 1 + x2 = t , then 2xdx = dt
  In the second part say arctan(x) = u , then du = dx / (1+x2)
  
     /  dt      /
  =  | ---  +   | u du
     /  t       /
  
                 2                     2
               u           2     arctan (x)
  =   ln|t| + ---- = ln|1+x  | + ----------- + C
                2                    2
  

• 
  

  / ax | e | --------dx | ax / e + b

  
  

   
  say eax + b  =  u then  aeaxdx = du
  
    1  /  du     1           1      ax
  = -  | ---  =  - ln|u| =  --- ln|e   + b| + C
    a  /  u      a           a
  
  

• 
  

  / I= | (x + 3) cos(x) dx /

  
  

   
  say (x + 3) = u and cos(x) dx =  dv
  Integration by parts gives
  
                       /
  I = (x + 3) sin(x) - |  sin(x) dx
                       /
  
    = (x + 3) sin(x) + cos(x) + C
  
  

• 
  

  / | 2 + x | -------- dx | 2 - x /

  
  

   
         /                    /                    /
         |  x + 2             |  x -2 + 4          |         4
   = -   | -------- dx  = -   | --------- dx = -   | (1 + -------) dx
         |  x - 2             |  x - 2             |       x - 2
         /                    /                    /
  
   = -x -4 ln |x - 2| + C
  
  

• 
  

  / | x arcsin(x) | ----------- dx | ________ | | 2 | \| 1 - x /

  
  

   
  
  say  arcsin(x) = u and the rest  =  dv
  
  then
               /                              ________
               |    x                         |      2
          v =  | ----------- dx  = ... =  -  \| 1 - x
               |    ________
               |    |      2
               |   \| 1 - x
               /
  
  Integration  by parts gives
                         ________    /   ________
                         |      2    |   |      2         1
          = - arcsin(x) \| 1 - x  +  |  \| 1 - x   . --------------dx
                                     |                  ________
                                     /                  |      2
                                                       \| 1 - x
  
                         ________
                         |      2
          = - arcsin(x) \| 1 - x   + x + C
  
  

• 
  

  / | |ex . cos( 1 + ex) dx | /

  
  

   
    say    1 + ex = u then  exdx  = du
  
  
     /
  =  | cos(u) du = sin u  + C  = sin ( 1 + e )x + C
     /
  
  

Level 3

• 
  

  Solve : /x | sin(2t + pi/2) dt = sqrt(3)/4 /0

  
  

   
  <=>  -(1/2) cos( 2x + pi/2) = sqrt(3)/4
  
  <=>  (1/2) sin(2x) = sqrt(3)/4
  
  <=>  sin(2x) =  sqrt(3)/2
  
  <=>  sin(2x) = sin (pi/3)
  
  <=>  2 x = pi/3 + 2 k pi  ; 2x = 2 pi/3 + 2k pi
  
  <=>  x = pi/6 + k pi ; x = pi/3 + kpi
  

• 
  

  Find ________ /5 | 2 | \| x - 9 I = | ---------- dx | x /3

  
  

   
  
  First we calculate a primitive function
           ________              ________
       /   |  2              /   |  2
       |  \| x  - 9          | x\| x  - 9
       |  ---------- dx  =   |  ---------- dx
       |      x              |       2
      /                     /       x
  
  say  x2 - 9 = u2 , then  x dx = u du
  
     /                  /   2                   /
     |   u.u du         |  u  + 9 -9            |   du
   = | ----------  =  = | ---------- du = u - 9 | ----------
     |   2              |   2                   |   2
     /  u  + 9          /  u  + 9               /  u  + 9
  
  
  
   = u - 3 arctan(u/3)
                              ________
       ________               |  2
       |  2                  \| x  - 9
   =  \| x  - 9  - 3 arctan ----------
                                3
  
  
                                        ________ |5
                ________               |  2      |
                |  2                  \| x  - 9  |                  4
  Thus, I =    \| x  - 9  - 3 arctan ----------  |  = 4 - 3 arctan(---) = 1.218
                                         3       |                  3
                                                 |3
  
  

• 
  

  Find /pi/2 | 1 - cos(u) g(x) = | -------------- du | sin2 (u) / x

  
  

   
  Let  2t = u  ; then  du = 2 dt   and first we'll calculate
  
     /
     |  1 - cos(2t)
   2 |-------------- dt  =
     |  sin2 (2t)
    /
  
  With carnot we write 1 - cos(2t) = 2 sin2(t)
  and sin(2t) = 2 sin(t) cos(t) =>  sin2 (2t) = 4 sin2(t) cos2(t)
  So, the indefinite integral is equal to
  
     /
     |  2 sin2(t)
   2 |----------------------- dt  =
     | 4  sin2(t) cos2(t)
    /
  
     /
     |        1
     |---------------- dt  =  tan(t) + C = tan(u/2) + C
     |    cos2(t)
    /
  From this result, we write
  
                   |pi/2
  g(x) =  tan(u/2) |       = tan(pi/4) - tan(x/2)
                   |x
  

External links to integral calculation

Integration by parts

Partial fractions

Exponential functions

Trigonometric functions

Irrational functions

All kinds of integrals