calculus(integration)

Introduction -

Most of the students find it really hard to deal with calculus. Here I had attempted to make it as simple as possible and keep all those tedious facts aside that are not important from the examination point of view. If something is still missing than there are always comments to let your opinion heard. The topic that we are going to study here is “INDEFINITE INTEGRALS”.

What is integration?  -

Integration in simple words is the reverse process of differentiation i.e. a differentiated equation (over which we have performed the task of differentiation) when undergoes the process of integration, yields the original equation as it was prior to applying differentiation. For e.g;

Let we have an equation say we have f(x) =2x² + 4x                 -eq. (1)

 --- { Here f(x) means a function of ‘x’ i.e the equation contains “x” terms only }

So differentiating this eq. (1) we have “4x + 4” as its derivative or we can write that;

d/dx (2x² + 4x) = 4x + 4                                             - eq. (2)

Now if we perform integration over this eq.(2) we will definitely get 2x² + 4x which is nothing else but eq.(1) and hence we say that integration is the reverse process of differentiation or integration means Anti derivative.

                               

Symbol of integration -

While an equation to be differentiated is kept under brackets and put in front of the “d/dx” symbol

{ d/dx(2x² + 4x)}; the integration is shown using the symbol  “ ʃ “ and then keeping the equation to be integrated inside the brackets ahead of it;

ð  ʃ (4x + 4) dx = 2x² +4x

Types of integration –

Integration roughly means to cover a region/ occupy a region, according to the problem in hand. When we know the limits i.e the extremes with proper initial and final points inside which we have to occupy/calculate the area, then we say that we are dealing with the type of integration known as “definite integrals”. However when the initial or final points are unknown then we say that the query relates to “indefinite integrals”. Indefinite integrals are the generalized form and providing specific limits over them converts the integral into definite integrals. We will limit our discussion only to indefinite integrals here.

Fig: Shows structure of definite indefinite integral

Fig: Shows structure of definite integral

How is integration done?/ Methods to perform integration-

It becomes extremely cumbersome for us to each and every time follow the inverse process of differentiation and find out the required solution, thus there is a need of definite formulae that help us accelerate and reduce the overload of solving such equations that are nearly impossible to solve with the anti-derivative way (finding integration in the reverse order of differentiation). It is advised to go through the formulae as thoroughly as possible so that you could pace up and finish the requirements of the problem as quickly as possible. Another very important thing to keep in mind is that while dealing with definite integrals we may reach to a solution that is not similar to the one we get solving the similar problem some other time by a different method. Note that it absolutely does not mean that you are wrong. What matters here is that the correct method has been employed (by correct method we mean that proper formulae have been applied and the equation has been reduced properly), hence many different but correct solution may exist for a similar problem and we can, in some way or the other reach on one correct answer from the other correct answer as obvious. One must note that in definite integrals with constant limits we have a single definite answer that is unique.

Before taking up the various integration formulae let us see with the help of an example that how could we integrate following reverse differentiation in detail.

Query:-  To calculate integration/primitive of 4x³ + 5x⁴

Sol:- Hence we look for a function whose derivative is 4x³ + 5x⁴. Note that

d/dx(x⁴ + x⁵) = 4x³ + 5x⁴

hence it is sure that the integration of 4x³ + 5x⁴ is x⁴ +x⁵.

This is also known as hit and trial method or method of inspection.

IMPORTANT NOTE :- Until now we have discussed integration as reverse of differentiation but it should be noted that we must add a “constant” to our solution after the integration has been performed on an equation/function. This is due to the following fact-

d/dx(3x² + 4x³) = 6x + 12x²                                                                           -- eq.(3)

but also;

d/dx(3x² + 4x³ +4) = 6x + 12x²                            --{ differentiation of constant is zero}    eq.(4)

Note that both the above equations are different but have the same answer after being differentiated. Now as we have studied till now that the integration is reverse of differentiation so when integrated both eq.(3) and eq.(4) must yield 3x² + 4x³ , which is true for eq.(3) but not with eq.(4) , hence there arises a need to add a symbol depicting a constant(since only the differentiation of a constant is zero) in front of the solution equation after the integration process that depicts the possibility of existence of a constant which might had been eliminated because of the differentiation process with the equation that has just been integrated. In this discussion we will represent the integration constant with the term “c” or “c₁” , “c₂” etc. Clearly c = 0 in eq.(3) while c = 4 in eq.(4) .

Coming back to the discussion let us now go through the various integration formulae that will be very helpful in solving the problems. We will study each of this in detail with individual formulae based problems. These are summarized as under;

Integration formulae:-  

   Derivatives                                                      Integrals

  d/dx (x) = 1 ;                                                    ʃ dx = x + c    --{“x” is any variable}

  generally;                                                         generally;

 d/dx(x ^{n + 1})/(n + 1)=xⁿ ;                                                ʃ xⁿ dx = x^{n +1}/(n + 1) + c ;{important}

2.> d/dx (sin x) = cos x ;                                                ʃ cos x dx = sinx ;

3.> d/dx (- cos x) = sin x ;                              ʃ sin x dx = - cos x + c ;

4.> d/dx (tan x) = sec²x ;                               ʃ sec² x dx = tan x + c ;

5.> d/dx (- cot x) = cosec²x  ;                       ʃ cosec² x dx = -cot x + c ;

6.> d/dx (sec x) = (sec x).(tanx)  ;              ʃ (sec x)(tan x) dx = sec x +c ;

7.> d/dx ( - cosec x) = (cosec x).(cot x) ;                 ʃ (cosec x)( cot x) dx = - cosec x + c ;

8.> d/dx (sin^-1 x) = 1/√ ( 1 – x²) ;                                ʃ dx / √ ( 1 – x²) = sin^-1 x + c ;

9.>d/dx ( - cos^-1 x) = 1/√ ( 1 – x²) ;             ʃ dx / √ ( 1 – x²) =  - cos^-1 x + c ;

10.> d/dx ( tan^-1 x) = 1/(1 + x²) ;                 ʃ dx / (1 + x²) = tan^-1 x + c ;

11.> d/dx (- cot^-1 x) = 1/(1 + x²) ;                ʃ dx / (1 + x²) = - cot^-1 x + c ;

12.> d/dx ( sec^-1 x) = 1 / x . √ (x² -1) ;        ʃ dx / x . √(x² -1) =  sec^-1 x + c ;

13.> d/dx (- cosec^-1 x) = 1 / x . √(x² -1) ;   ʃ dx / x . √(x² -1) = - cosec^-1 x + c ;

14.> d/dx (e^x) = e^x ;                                         ʃ e^x  dx = e^x + c ;

15.> d/dx (log |x|) = 1/x ;                                            ʃ ( 1/x)  dx =  log |x| + c ;

16.> d/dx ( a^x/log a) = a^x ;                                             ʃ a^x dx = a^x / log a + c ;     { “a” is any constant}

Similar to the fact that d/dx is known as derivative w.r.t “x”; we would say that ʃ (Any function)dx  means integration of that function w.r.t “x”. Note that functions such as e^x, log |x|, cos x ,x⁴ + 5x + 7 etc are all functions of “x” as they include terms including “x”. It is clearly observed that in each formulae integrals are opposite to their corresponding differentials. We note that “c” is the constant of integration above as discussed earlier. Let us move forward and take some questions based on each formulae step by step. The remaining article depicts how different type of questions are solved by firstly evaluating the type of formulae (see 1 to 16 above) applicable to the problem and secondly the method to do this. Before moving forward make sure that you have understood all what has been discussed above. Assuming each point till now is clear to you let us now move deep into the topic and make its study a fun and interesting thing.

TYPE 1:- Questions based upon  ʃ xⁿ dx = x^{n +1}/(n + 1) + c .

Example 1.>  ʃ  (x³ – 1)/ x²  dx

Solution.> We have

ʃ (x³ – 1)/x²  dx = ʃ x dx - ʃ x^-2 dx        -- {simply dividing the numerator by denominator}

                                                                                --{Note that we can separate the terms of an equation to be integrated into two or more different integral equations like done above as in case of simple polynomial equations}

= { ( x¹⁺¹/ 1+ 1 ) + c1 }  -  { ( x^{-2 + 1}/-2 + 1 ) + c2         --{applying the formulae}

--{ c1 and c2 are the constants of integration for each of the two integrals }

= x²/2 + c1 – x^-1/-1 –c2

= x²/2 + 1/x + c1 – c2

=x²/2 +  1/x + c                                                 

--{ Here c  = c1 - c2 (say) ; this is in order to reduce the number of constants as the difference of two constants is always a constant }

This is the required solution.

Example 2.> ʃ (x^2/3 + 1 ) dx

Solution 2.> We have ʃ (x^2/3 + 1 ) dx

=  ʃx^2/3 dx + ʃ dx

= x^{2/3 + 1}/(2/3 + 1) + x + c

= (3/5) x^5/3 + x + c             which is the required solution.

TYPE 2:- Questions based upon ʃ cos x dx = sinx ;

Example- ʃ  (3/4) cos x dx

Solution:- ʃ (3/4) cos x dx

= (3/4) ʃ cos x dx                                    

-{Remember that constants when in product with a term should be taken out of the bracket since they have no effect over integration computation}

= (3/4) sin x + c    -(Solution)

TYPE 3:- Questions based upon ʃ sin x dx = - cos x + c ;

Example:-  ʃ ( 1/ cosec x ) dx

Solution:- =  ʃ sinx dx                      --{ 1/ cosec x = sin x}

= - cos x + c         -(Solution)

TYPE 4:- Questions based upon  ʃ sec² x dx = tan x + c ;

Example :-  ʃ tan² x dx

Solution:- ʃ tan² x = ʃ ( sec² x – 1 ) dx                        --{tan 2  x = sec 2 x - 1 }

= ʃ sec ² x dx - ʃ (1) dx                     

=  tan x – x + c                    - -{constants have been combined and assumed to be equal to “c”}

TYPE5 :- Questions based upon  ʃ cosec² x dx = -cot x + c ;

Example :-  ʃ (cos 2x) ÷ (sin ² x)(cos ² x) dx

Solution :-  ʃ (cos 2x) ÷ (sin ² x)(cos ² x) dx

= ʃ ( cos ² x – sin ² x ) /( sin²  x )( cos ² x)   dx          --{ since cos 2x = cos ² x – sin ² x }

= ʃ 1 / sin ² x dx - ʃ 1 / cos ² x dx                    --{ Devide numerator by denominator }

= ʃ cosec ² x dx - ʃ sec ² x dx

=  – cot x – tan x + c (answer)

TYPE6 :- Questions based upon   ʃ (sec x)(tan x)dx = sec x +c ;

Example :- ʃ  1 / ( 1 + sin x) dx

Solution :-  ʃ  ( 1 / 1 + sin x)  dx

=  ʃ  ( 1 / 1 + sin x)  × ( 1 – sin x ) / ( 1 – sin x) dx                    --{ rationalising – look below for meaning}

= ʃ  ( 1 – sin x ) / ( 1 – sin ² x) dx  

= ʃ  ( 1 – sin x ) / cos ² x dx                                             --{ 1 – sin ² x = cos ² x } as  { sin² x+ cos² x = 1 }

= ʃ  ( 1 / cos ² x)  dx  -  ʃ  (sin x / cos ² x ) dx             

--{ separating into two different integrals as done before }

= ʃ  sec ² x dx  – ʃ  sin x / ( cos x )( cos x) dx            

= ʃ  (sec ² x  dx    –  ʃ (tan x ) ( sec x )  dx                   --{ sin x / cos x = tan x and 1 / cos x = sec x }

= tan x – sec x + c             ( solution).

Remark :-

Ø  By the term/ technique rationalizing( in math ) we mean to multiply and divide the given term with the denominator of itself with the sign between both the terms of the denominator changed i.e positive to negative and negative to positive. Note that the terms should be “two” exactly (in the denominator of the given term) with either positive or negative sign between them to be able to get rationalize.  Here the first of the two terms is “1” and the second term is “sin x” with +ve sign between them. Change positive to negative and then multiply-divide with the original term to obtain the solution as given.

Ø  It is worth noticing here that we use techniques to change the given terms into some specific terms all the times coz we have the integration defined for only the later terms and so the expression had to be reduced anyhow into these known integrals. In future we will see more of the methods on how to change the integrals and reduce them to solutions by other methods. Hence methods with which the given examples are solved are not the only methods that apply to each type class and continuous practice can take the best out of the student to deal with such problems.

TYPE7 :- Questions based upon  ʃ (cosec x)( cot x) dx = - cosec x + c ;

Example :-  ʃ ( 2 + 3cos x) / sin ² x dx

Solution :-  ʃ  2 / sin ² x dx + ʃ  3cos x / sin ² x dx

=  ʃ ( 2cosec ² x )dx + ʃ (3cot x cosec x)dx                                --{ cos x / sin x = cot x and 1 / sin x = cosec x }

=  -2cot x -3cosec x + c (Required solution)

TYPE-8 TO TYPE-13 :-  :- Questions based upon 

Ø ʃ dx / √ ( 1 – x²) = sin^-1 x + c ; 

Ø ʃ dx / √ ( 1 – x²) =  - cos^-1 x + c ; 

Ø ʃ dx / (1 + x²) = tan^-1 x + c ;

Ø  ʃ dx / (1 + x²) = - cot^-1 x + c ;

Ø  ʃ dx / x.(√x² -1) =  sec^-1 x + c ; 

Ø ʃ dx / x.(√x² -1) = - cosec^-1 x + c ;

The questions based on above problems could be divided mainly into two parts.

1 .> Direct questions:-

Example :- ʃ  1 / √ { 1 – (sin x )² }  dx

= sin^-1 (sinx)  dx                

--{  Here x = sin x ; this is possible since the formulae is valid for every term and not only “x” with power “2”. The fact apply everywhere. }

= x          (solution)                            { since sin^-1( sin x )dx  =  x}

2.> Reducible type questions :-

Example :- ʃ  tan ^-1 { √ ( 1 – cos 2x ) / √ (  1 + cos 2x ) } dx

Solution :- tan ^-1 { √ 2sin² x / √2cos ² x} dx

= ʃ tan ^-1 (tan x) dx                          

--{ cancelling “2” from numerator and denominator and taking tan² x out of the root as tan x}

= x          (Solution)

TYPE 14 AND TYPE 16:- Let us take a mixed example here based on ʃ e^x  dx = e^x + c ; and

ʃ ( 1/x)  dx =  log |x| + c  :-

Example :- ʃ  ( 3e^x + 1/x ) dx

= 3 ʃ e^x dx + ʃ 1/x dx

= 3e^x + log x + c                 (Solution)

Remarks:  What all we studied above was what we can call as stepping in into the integral topic. If you are through with what we studied than you are ready for the various methods that bear extreme importance from the examination point of view in this chapter. Check back again to see when I post them.  You could find the complete list of trigonometric formulae in the coming post from MathKnackChannel . And yes in the mean time  you could try some questions for your practice.

Ø  (cx + d )² dx

Ø  (4e^5x + 1) dx

Ø  (√x – 1 / √x )² dx

Ø  1 / ( 1 + cos x ) dx

Ø  Sec x / ( sec x + tan x) dx

Ø  ( Cos 2y – cos 2a ) / ( cos y – cos a ) dy                             --{ w.r.t “y”}

Ø  Sin( e^x ) d( e^x)                                             --{w.r.t “ e^x ”}

Ø  If f’( x ) = 1/x + 1 / ( 1 + x² )  dx and given that f( 1 ) = π/4 , than find f( x ). Where f’( x ) is differentiation of the function f( x )  w.r.t “x” .

HINT :- f( x ) = ʃ f ' ( x ) dx                               

--{ integration is the reverse process of differentiation and when they are applied at the same time over a function both of them cancel each other and the function remains unchanged }

That’s it from my side in this post. So keep practicing. KEEP INTEGRATING. Lol ! Thanks :)