tg

tg
tgt

Monday, 4 August 2014

CHAPTER 1 - Integration - Indefinite Integration

BASIC RULES AND FORMULAE

The following is a set of straight forward rules pertaining to integration, that follow by definition:
(a) A constant is always included in the expression for the indefinite integral, i.e.,
if
g'\left( x \right) = f\left( x \right), then
\int {f\left( x \right)dx = g\left( x \right) + C}
This is because, as mentioned earlier, the derivative of a constant is 0.
(b) The integral of a derivative gives back the same function itself (with a constant):
\int {f'\left( x \right)dx = f\left( x \right) + C}
The derivative of an integral also gives the same function:
\dfrac{d}{{dx}}\left( {\int {f\left( x \right)dx} } \right) = f\left( x \right)
These two results are in agreement with the fact that differentiation and integration are inverse operations.
(c) \int {\left\{ {f\left( x \right) \pm g\left( x \right)} \right\}} \,dx = \,\,\int {f\left( x \right)dx + \int {g\left( x \right)dx} }
(d) \int {k\,f\left( x \right)dx\,\, = \,\,k\int {f\left( x \right)dx} }
(e) if \int {f\left( x \right)dx\,\, = \,g\left( x \right) + C,}  then
\int {f\left( {ax + b} \right)dx\,\, = \dfrac{1}{a}\,g\left( {ax + b} \right) + C} \ldots(i)
How is this true? Since g(x) is the anti-derivative of f(x),g'(x) = f(x).
Now we differentiate (i) :
f\left( {ax + b} \right) = \dfrac{1}{a}\dfrac{d}{{dx}}\left( {g\left( {ax + b} \right)} \right)
 = \dfrac{1}{a} \cdot g'\left( {ax + b} \right) \cdot a
 = g'\left( {ax + b} \right)
 = f\left( {ax + b} \right)
This shows that (i) holds true.
This result is quite useful as we’ll realise in the course of studying this chapter.
We now present a table of some basic integration formulae. You are urged to verify the truth of these formulae by differentiating the right hand side of each formula and check whether the expression you obtain is equal to the one inside the integral on the left hand side, or not:

BASIC INTEGRATION FORMULAE

1. \int {{x^n}dx}  = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C\,\,\,\,\,;\,\,\,\,\,n \ne  - 1*11. \int {\cot xdx = \ln \left| {\sin x} \right| + C}
2. \int {\dfrac{1}{x}} \,dx = \ln x + C12. \int {\tan xdx =  - \ln \left| {\cos x} \right| + C}
3. \int {{e^x}} dx = {e^x} + C13. \int {\sec xdx = \ln \left| {\sec x + \tan x} \right| + C}
4. \int {{a^x}} dx = \dfrac{{{a^x}}}{{\ln a}} + C14. \int {{\rm{cosec}}\,xdx = \ln \left| {{\rm{cosec}}\,x - \cot x} \right| + C}
5. \int {\sin \,xdx =  - \cos x + C} 15. \int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}dx}  = {\sin ^{ - 1}}\dfrac{x}{a} + C
6. \int {\cos \,xdx = \sin x + C} 16. \int {\dfrac{{ - 1}}{{\sqrt {{a^2} - {x^2}} }}dx}  =  - {\sin ^{ - 1}}\dfrac{x}{a} + {C_1}

= {\cos ^{ - 1}}\dfrac{x}{a} + {C_2}
7. \int {{{\sec }^2}xdx = \tan x + C} 17. \int {\dfrac{1}{{{a^2} + {x^2}}}dx}  = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C
8. \int {{{{\mathop{\rm cosec}\nolimits} }^2}xdx =  - \cot x + C} 18. \int {\dfrac{{ - 1}}{{{a^2} + {x^2}}}dx}  = \dfrac{{ - 1}}{a}{\tan ^{ - 1}}\dfrac{x}{a} + {C_1}
 = \dfrac{1}{a}{\cot ^{ - 1}}\dfrac{x}{a} + {C_2}
9. \int {\sec x\tan xdx = \sec x + C} 19. \int {\dfrac{1}{{x\sqrt {{x^2} - {a^2}} }}dx}  = \dfrac{1}{a}{\sec ^{ - 1}}\left( {\dfrac{x}{a}} \right) + C
10. \int {{\mathop{\rm cosec}\nolimits} x\cot xdx =  - {\rm{cosec}}\,x + C} 20. \int {\dfrac{{ - 1}}{{x\sqrt {{x^2} - {a^2}} }}dx}  = \dfrac{{ - 1}}{a}{\sec ^{ - 1}}\dfrac{x}{a} + {C_1}
 = \dfrac{1}{a}{{\mathop{\rm cosec}\nolimits} ^{ - 1}}\dfrac{x}{a} + {C_2}
* In particular, \int {dx = x + C}
For example, to prove that formula (17) is true, we differentiate the right side:
\dfrac{d}{{dx}}\left( {\dfrac{1}{a}{{\tan }^{ - 1}}\dfrac{x}{a}} \right) = \dfrac{1}{a}\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{a}} \right)} \right)
 = \dfrac{1}{a}\dfrac{1}{{1 + \dfrac{{{x^2}}}{{{a^2}}}}} \cdot \dfrac{1}{a}
 = \dfrac{1}{{{a^2}}} \cdot \dfrac{{{a^2}}}{{{x^2} + {a^2}}}
 = \dfrac{1}{{{x^2} + {a^2}}}
Therefore,
\int {\dfrac{1}{{{x^2} + {a^2}}}dx}  = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + C
Suppose that you now have to evaluate the following integral:
\int {\dfrac{1}{{{a^2} + {{\left( {bx + c} \right)}^2}}}dx}
We will use formula (17) and rule (e) stated earlier:
\int {\dfrac{1}{{{a^2} + {{\left( {bx + c} \right)}^2}}}dx}  = \dfrac{1}{a}{\tan ^{ - 1}}\left( {\dfrac{{bx + c}}{a}} \right) \times \dfrac{1}{b} + C
 = \dfrac{1}{{ab}}{\tan ^{ - 1}}\left( {\dfrac{{bx + c}}{a}} \right) + C
Observe carefully how we obtained the final expression.
There is a variety of methods in which we can evaluate indefinite integrals. We can broadly divide these methods into five major categories:
(1) SIMPLE REARRANGEMENTS :
We rearrange the given expression in such a way so that we obtain a combination of the basic integrals that we have just discussed.
(2) SUBSTITUTIONS :
We use some substitution to convert the given expression into a more conveniently “integrable” form.
(3) EXPANSION USING PARTIAL REACTION :
This method is applicable to rational algebraic functions; we use a partial dfractions expansion to split such a function into more elementary functions that can easily be integrated.
(4) INTEGRATION BY PARTS :
This powerful method can be applied to the product of any two arbitrary functions.
(5) REDUCTION FORMULAE :
These formulae make it possible to reduce an integral depending on the index n > 0, called the order of the integral, to an integral of the same type but with a smaller index.
All these methods will now be discussed in detail.
A word of advice: make it a point to practice as much questions as possible for integration; only then can you get the ‘hang’ of it. You should even attempt the solved examples on your own before looking at the solutions.
Post a Comment