Basic Integral formula

                      

                                  

BASIC  INTEGRAL FORMULA

                                                             

 Integral Formulas

    ∫ 1 dx = x + C

    ∫ a dx = ax+ C

    ∫ xn dx = ((xn+1)/(n+1))+C ; n≠1

    ∫ (1/x) dx = ln |x| + C

    ∫ ex dx = ex+ C

    ∫ ax dx = (ax/ln a) + C ; a>0,  a≠1

Trigonometric integration formula

   ∫ sin x dx = – cos x + C

   ∫ cos x dx = sin x + C

   ∫ sec2x dx = tan x + C

   ∫ csc2x dx = -cot x + C

   ∫ sec x (tan x) dx = sec x + C

   ∫ cosec x ( cot x) dx = – cosec x + c

   ∫ tan x = log |sec x| + C

   ∫ cot x = log |sin x| + c 

   ∫ sec x = log |sec x + tan x| + C

   ∫ cosec x = log |cosec x - cot x| + C

Inverse trigonometric integration formula

   ∫1/√(1 - x²) dx = sin^-1x + C

   ∫ 1/√(1 - x²) dx = -cos^-1x + C

   ∫1/(1 + x²) dx = tan^-1x + C

   ∫ 1/(1 + x² ) dx = -cot^-1x + C

   ∫ 1/x√(x² - 1) dx = sec^-1x + C

   ∫ 1/x√(x² - 1) dx = -cosec^-1 x + C

Some important integral formula

   ∫1/(x² - a²) dx = 1/2a log|(x - a)(x + a| + C

   ∫ 1/(a² - x²) dx =1/2a log|(a + x)(a - x)| + C

   ∫1/(x² + a²) dx = 1/a tan^-1x/a + C

   ∫1/√(x² - a²)dx = log |x +√(x² - a²)| + C

   ∫ √(x² - a²) dx = x/2 √(x² - a²) -a²/2 log |x + √(x² - a²)| + C

   ∫1/√(a² - x²) dx = sin^-1 x/a + C

   ∫√(a² - x²) dx = x/2 √(a² - x²) dx + a²/2 sin^-1 x/a + C

   ∫1/√(x² + a² ) dx = log |x + √(x² + a²)| + C

   ∫ √(x² + a² ) dx = x/2 √(x² + a² )+ a²/2 log |x + √(x² + a²)| + C

Integration by parts formula

∫ f(x) g(x) dx = f(x) ∫g(x) dx - ∫ (∫f'(x) g(x) dx) dx + C

Integration by substitution formula

 If I = ∫ f(x) dx, where x = g(t) so that dx/dt = g'(t), then dx = g'(t)

               and          I = ∫ f(x) dx = ∫ f(g(t)) g'(t) dt

INTEGRATION

Integration is the process of combining or adding parts to make a whole. It can be 

used for finding the areas ,volumes and many useful things. Integration, in math is 

 the method of finding a function g(x) the derivative of which, d/dx g(x), is equal to 

a given function f(x). This is indicated by the  sign “∫f(x),” integral as  ∫, usually 

called the indefinite integral of the function. The symbol dx represents an 

infinitesimal displacement along x ; thus    ∫ f(x)dx is the summation of the product 

of f(x) and dx. There are three methods of integration

    

     (i)Integration by substitution

     (ii) Integration by  parts

     (iii)  Integration by partially

(i)  Integration by substitution method ;

Integration by substitution method is also known as u-substitution, reverse chain 

rule or change of variables, is a method for evaluating integrals and antiderivatives. 

 Integration by substitution formula is given by

 If I = ∫ f(x) dx, where x = g(t) so that dx/dt = g'(t), then dx = g'(t)

               and          I = ∫ f(x) dx = ∫ f(g(t)) g'(t) dt

Example ; Compute   ∫ (5x +2)⁴ dx

solution

       Let u = (5x +2)

      

        du = 5  dx 

      

        Thus ∫ (5x +2)⁴ dx = 1/5∫(u)⁴ du               

                            

                                     = 1/5. u⁵ /5

                                   

                                          = u⁵ /25

                     

                                          = (3x+2)⁵ /25

(ii)  Integration by parts formula

It is a method of integration when two functions are multiplied together

∫ f(x) g(x) dx = f(x) ∫g(x) dx - ∫ (∫f'(x) g(x) dx) dx + C

more precisely,

                             ∫u dv=u v−∫v du

Example ; ∫x cos x dx 

solution

     

       u = x and v= cos x

    

     du = dx and dv = sin x

    ∫x cos x dx =  ∫u v dv

                  

                    = u v - ∫ v du

                 

                   = x sin x   -  ∫ sin x dx

                  

                   =  x sin x + cos x + c

(iii)  partial Integration

 

 Partial fractions are the sum of proper rational functions  when we decompose an 

improper rational function. The improper rational function means the rational 

function with the degree of its numerator  which is not less than the degree of the 

denominator. Formulas to decompose the improper rational functions to get partial 

fractions are given below;

                             

 Example;  Find ∫ dx / [(x + 1) (x + 2)]

 solution;

    

        1 / [(x + 1) (x + 2)] = A / (x + 1) + B / (x + 2) … (1)

                      

              Solving this equation, we get,
                        A (x + 2) + B (x + 1) = 1 

                       Or, Ax + 2A + Bx + B = 1

                        x (A + B) + (2A + B) = 1

              To make  LHS and  RHS to be equal, 

             we have ,A + B = 0 and 2A + B = 1.

            On solving these two equations, we get

             A = 1 and B = – 1.

                   Therefore, 1 / [(x + 1) (x + 2)] = 1 / (x + 1) – 1 / (x + 2)

                    Hence, ∫ dx / [(x + 1) (x + 2)] = ∫ dx / (x + 1) – ∫ dx / (x + 2)

                    = log |x + 1| – log |x + 2| + C         

Some problems using integration formula

1. 

solution

     

                    
 

                                  

                           

                                 

                                 

    2. 
     

solution

           

            
         

            

            

              

 

                                

3.  ∫tan8x sec4 x dx

  Solution:

    

           Given: ∫tan8x sec4 x dx

   

               Let I = ∫tan8x sec4 x dx — (1)

   

               sec4x = (sec2x) (sec2x)

     

            Now, substitute in (1)

   

                   I = ∫tan8x (sec2x) (sec2x) dx

                     = ∫tan8x (tan2 x +1) (sec2x) dx

      

                     = ∫tan10x sec2 x dx + ∫tan8x sec2 x dx

                    =( tan11 x /11) + ( tan9 x /9) + C

          Hence, ∫tan8x sec4 x dx = ( tan11 x /11) + ( tan9 x /9) + C

4.  Write the anti-derivative of the following function: 3x2+4x3

   Solution:

              Given: 3x2+4x3

              The antiderivative of the given function is written as:

 

              ∫3x2+4x3 dx =  3(x3/3) + 4(x4/4)

              

                                 = x3 + x4

 

              Thus, the antiderivative of 3x2+4x3 = x3 + x4

5. Integrate  2x sin(x2+ 1) with respect to x:by substitution method

   

  Solution:

      

              Given function: 2x sin(x2+ 1)

                      d/dx ( x2 + 1) =  2x.

              By substitution method, we get

                 x2 + 1 = t, so that 2x dx = dt.

   

                Hence, we get ∫ 2x sin ( x2 +1) dx = ∫ sin t dt

                                                                     = – cos t + C

                                                                      = – cos (x2 + 1) + C

              Therefore,  ∫  2x sin(x2+ 1) = – cos (x2 + 1) + C by substitution method

6.  Integrate: ∫ sin3 x cos2x dx

   Solution:

 

                 Given that, ∫ sin3 x cos2x dx

 

                 ∫ sin3 x cos2x dx = ∫ sin2 x cos2x (sin x) dx

                                           =∫(1 – cos2x ) cos2x (sin x) dx —(1)

   

                 substitute t = cos x,

                  Then dt = -sin x dx

 

                 ∫ sin3 x cos2x dx = – ∫ (1-t2)t2 dt

                                            = – ∫ (t2-t4) dt  (by multiplying t2 

                                             = – [(t3/3) – (t5/5)] + C —(2)

                  Now, substitute t = cos x in (2)

                                             = -(⅓)cos3x +(1/5)cos5x + C

                   Hence, ∫ sin3 x cos2x dx = -(⅓)cos3x +(1/5)cos5x + C

7.  Prove 

solution

                 
    

                           

                          

                         

8. 

Solution

                    

                      

                     

                   

9. 

 Solution

                   

                       
 

                        

10. 

Solution