Calculas seem to be latin to you ? So here is the way out to make calculas your strength!
This topic covered a brief discussion on Integration - an important branch of calculas. From the historical stand point Integration means summation of infinite series whose each term is infinitesimally small..Intgral calculus is the branch of mathematics we use to find lengths , areas , and volumes of irregular shapes.
Now we will define integration mathematically . Let f ( x ) be a given
function of x . If we can determine another function F ( x ) , such that its derivative with respect to x is equal to f ( x ) [ differential of F ( x ) is equal to f ( x ).dx . Then F ( x ) is
defined as an indefinite integral of f ( x ) .

It is not possible to shed light on the whole part of integration by a single article . So ,today let's start discussing on Integration by Parts method .It is the method of integration when we are to integrate product of two functions .

If u and v are two differentiable function of x , then ,

∫ uv d x = u ∫ dv - [ [ du / dx ∫ v dx ] dx
i.e . the integral of product of two functions = [ ( first function × integral of second function ) - ( integral of ( derivative of 1st function × integral of second function ]

But we have to remember the rule for selecting the order of choosing the function .We have to select the functions as u ( 1st function ) and v ( 2nd function ) according to the order of alphabate in the word ' I L A T E ' . The significance of the word I L A T E is described below :

I --denotes Inverse function

L – denotes Logarithmic function

A –denotes Algebraic function

T –denotes Trigonometric function

E –denotes Exponential function
Now let's try solving the following simple problems :

( 1 ) ∫ x e^ x dx = x ∫ e^ x dx - ∫ [d/dx ( x ) ∫ e^x dx ] dx
|Take u = x ( algebraic function ) |
| and v = e ^ x ( exponential function ) |

= x e^x - ∫ e ^ x dx | d/dx ( x ) = 1 |
= x e ^ x - e ^ x + C | C is arbitrary constant
= ( x – 1 ) e ^x + C
( 2 ) Find ∫ e ^ x Sin x dx

Let I = ∫Sin x e ^ x dx | Take u = Sin x ( trigonometric function )
| and v = e ^ x ( exponential function )
= Sin x ∫ e ^ x dx - ∫[ d /dx ( Sin x ) ∫ e ^ x ] dx

= Sin x e ^x - ∫ [Cos x e ^ x ] dx
| Again we have to apply By Parts to solve ∫Cosx e^x d x

= Sin x e ^ x - [ Cos x ∫ e ^ x - ∫ [d / dx ( Cos x ) ∫ e^ x ] dx

= Sin x e ^ x -[ Cos x e ^x - ∫ [ ( - Sin x ) e ^x ] dx

= e ^x ( Sin x – Cos x ) - ∫ Sin x e ^ x dx

I = e ^x ( Sin x – Cos x ) - I + C | Place ∫ Sin x e ^ x = I |

2 I = e ^x ( sin x – Cos x ) + C | C is arbitrary constant |

I = e ^x ( sin x – cos x ) / 2 + C

( 3 ) ∫ ( (lnx ) / ( 1 + ln x ) ² ) dx

= ∫( ( 1 + ln x - 1 ) / ( 1 + ln x ) ² dx

= ∫ ( dx / ( 1 + ln x ) ) - ∫ ( 1 / ( 1 + ln x ) ² ) ) dx | Apply integration by parts |

= ( 1 / ( 1 + ln x )) ∫dx - ∫ [ d / dx ( 1 / ( 1 + ln x ) ∫ dx ] dx
- ∫[ 1 / ( 1 + ln x ) ² ] dx + C
| C is arbitrary constant |

= x / ( 1 + ln x ) - ∫[ ( - 1 ) / ( 1 + ln x ) ² × ( 1 / x ) × ( x ) ] d x
- ∫ ( dx /( 1+ lnx ) ² + C

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

= x / ( 1 + ln x ) + C

Now try the following :

∫ Cos 2x ln ( 1 + tan x ) dx

After going through the above solved examples you might not find Integration by parts your night mare any more. If you still having some problem in solving problems help of an e tutor is suggested . A session on online integration would be of great help in case of practice solving online integration in an interactive way .