HOW TO LEARN INTEGRATION FORMULAE/FORMULAS USING TRICKS
Let us learn and remember most Important formulas of Integration , tips and tricks to learn algebraic ,most important differentiation questions for +2 maths, indefinite integration tricks and shortcuts trigonometric and by parts formulas in an easy and short cut manners.
Trigonometric Formulae
1. ∫ sin x dx = - cos x +c
The integration of sin x is - cos x ,then divide it with the derivative of its angle.
If we have to find the integration of sin 2x , then we shall find it as
Step 1. 1st find the integration of sin x which is - cos x .
Step 2. Divide it with the derivative of 2x ,which is 2, so
∫ sin 2x dx = - ( cos 2x) 2 + c ,
∫ sin 8x dx = - ( cos 8x) 8 + c ,
∫ sin 3x4 dx = - ( cos 3x4 ) 3 4 + c ,
Therefore ∫ sin nx dx = - ( cos nx) n +c ,
2. ∫ cos x dx = sin x + c
The integration of cos x is sin x ,then divide it with the derivative of its angle.
If we have to find the integration of cos 2x , then we shall find it as
Step 1 1st find the integration of cos x which is sin x .
Step 2 Divide it with the derivative of 2x ,which is 2, so
∫ cos 2x dx = ( sin 2x) 2 + c ,
∫ cos 8x dx = ( sin 8x) 8 + c ,
∫ cos 3x4 dx = ( cos 3x4 ) 3 4 + c ,
Therefore ∫ cos nx dx = ( sin nx) n +c ,
2. ∫ cos x dx = sin x + c
The integration of cos x is sin x ,then divide it with the derivative of its angle.
Step 1 1st find the integration of cos x which is sin x .
∫ cos 2x dx = ( sin 2x) 2 + c ,
∫ cos 8x dx = ( sin 8x) 8 + c ,
3 ∫ tan x dx = log |sec x| + c or - log |cos x| + c
The integration of tan x is log |sec x| + c or -log |cos x| + c , then divide it with the derivative of its angle.
If we want to find the integration of tan x² . The integration of tan x² is log |sec x² | + c , then divide it with the derivative of its angle.
Step 1
Find the integration of tan x² ,which is log |sec x² | or - log |cos x² |,
Step 2
Divide it with the derivative of angle x² ,which is 2x.
Therefore
∫ tan x² dx = -(1/2x) log |cos x² | + c or (1/2x)log |sec x² | + c
Step 2
Divide it with the derivative of angle x² ,which is 2x.
∫ tan x² dx = -(1/2x) log |cos x² | + c or (1/2x)log |sec x² | + c
4 ∫ cot x dx = log |sin x| + c or - log |cosec x| + c
The integration of cot x is -log |cosec x| +c , then divide it with the derivative of its angle.
If we want to find the integration of cot x² . Then integration of cot x² is -log |cosec x² | + c or log |sin x² | + c , then divide it with the derivative of its angle.
The integration of cot x is -log |cosec x| +c , then divide it with the derivative of its angle.
If we want to find the integration of cot x² . Then integration of cot x² is -log |cosec x² | + c or log |sin x² | + c , then divide it with the derivative of its angle.
Step 1
Find the integration of cot x² ,which is -log |cosec x² | or log |sin x² |,
Step 2
Divide it with the derivative of angle x² , which is 2x.
Therefore ∫ cot x² dx = -(1/2x) log |cosec x2 | +c or (1/2x ) log |sin x² | + c
5. ∫ sec x dx = log |sec x - tan x | + c
If we want to integrate sec√x .Then 1st of all we apply the formula of integration of sec(any angle) then divide with the formula of integration of √x, So we have
∫ sec√x dx = ( log |sec√x - tan √x | )(2√x) + c
If we want to integrate sec√x .Then 1st of all we apply the formula of integration of sec(any angle) then divide with the formula of integration of √x, So we have
∫ sec√x dx = ( log |sec√x - tan √x | )(2√x) + c
6 ∫ cosec x dx = - log |cosec x - cot x | + c
If we want to integrate cosec√x .Then 1st of all we apply the formula of integration of cosec (any angle) then formula of integration of √x, So we have
∫ cosec√x dx = - (log | cosec√x -cot √x | )(2√x) + c
7 ∫ sec² x dx = tan x + c
Because the derivative of tan x is sec² x , So the Anti derivative or Integration of sec² x is tan x .
∫ sec² √x dx = (2√x ) tan √x + c
∫ cosec² dx = - cot x + c
Because the derivative of cot x is - cosec² x , So the Anti derivative or Integration of cosec 2 x is - cot x .
8 ∫ sec x tan x dx = sec x +c
Because the derivative of sec x is sec x tan x ,Therefore the integration of tan x sec x is sec x .If we want to integrate sec√x .tan √x .Then its integration will be sec √x,
∫ sec √x tan √x dx = 2 √x sec √x + c
9 ∫ cosec x cot x dx = - cosec x + c
Because the derivative of cosec x is - cosec x cot x , Therefore the integration of tan x sec x is sec x .
∫ cosec √x cot √x dx = - ( 2 √x cosec √x ) +c
Integration of Trigonometric Functions
S N f(x)
∫dx
1
sin x
-cos x + c
2
cos x
sin x + c
3
tan x
ln|sec x| + c
4
cot x
ln|sin x| + c 5
sec x
ln|sec x + tan x |+ c
6
cosec x -ln|cosec x - cot x |+ c
S N | f(x) | ∫dx |
---|---|---|
1 | sin x | -cos x + c |
2 | cos x | sin x + c |
3 | tan x | ln|sec x| + c |
4 | cot x | ln|sin x| + c |
5 | sec x | ln|sec x + tan x |+ c |
6 | cosec x | -ln|cosec x - cot x |+ c |
Algebraic Formulae
1 ∫ (constant) dx = (constant ) x + c
Integration of constant function is the constant function itself multiplied by the variable .
∫ 5 dx = 5x +c
Integration of constant function is the constant function itself multiplied by the variable .
∫ 5 dx = 5x +c2 ∫ xn dx = xn+1 n+1dx + c ,
∫ x3 dx = x4 4 + c ,
To find the integration of function where variable "x" or f(x) has power 'n' , where "n" is any real number, we shall increase the power of "x" by 1 and divide it with increased power.
e.g
∫ x² dx = {1/(2+1)} x2+1 + c
∫ (x ) ⅔ dx = (x ) (⅔)+1 (⅔)+1 + c ,
= 3(x ) 5/3 5 + c ,
∫ (ax+b ) n dx = (ax+b ) n+1 a(n+1) + c ,
∫ (3x + 7 )² dx = (3x+7 )2+1 3(2+1) + c ,
= (3x+7 )³ 9 + c
If we have to integrate sum of two functions ,then we shall integrate it separately as follows
To find the integration of function where variable "x" or f(x) has power 'n' , where "n" is any real number, we shall increase the power of "x" by 1 and divide it with increased power.
e.g
∫ (x ) ⅔ dx = (x ) (⅔)+1 (⅔)+1 + c ,
= (3x+7 )³ 9 + c
If we have to integrate sum of two functions ,then we shall integrate it separately as follows
4 ∫ [ f(x) + g(x)] dx = ∫f(x)dx + ∫ g(x) dx + c
∫ [{ x²}³ + (2x) ]dx = ∫ { x²}³ dx + ∫ (2x) dx
=∫ x⁶dx + ∫ 2x dx =
x 6+1 6+1 +(2/2)x² + c
= x⁷ 7 + x² + c
∫ {4x² + 3x }dx = ∫4x² + ∫ 3x dx
= 4×(1/3)x³ + (3/2)x² + c
=∫ x⁶dx + ∫ 2x dx =
= x⁷ 7 + x² + c
∫ [ 6x / 3x²] dx = log |3x² | + c
Memorize these integration formulas along with differentiation in Hindi
Integration By Parts
∫ [ f(x) g(x)] dx = f(x) ∫ g(x) dx - ∫ {f '(x) ∫ g(x) dx}dx + c
∫ x sin x dx = x ∫ sin x dx - ∫x' { ∫ sin x dx}dx + c
= x(-cos x) - ∫ (-cos x)dx + c
= -x cos x - sin x +c
∫ log x dx = ∫ log x.1 dx
= log x ∫x - ∫f '(log x) ( x )dx + c
= log x .1 - ∫(1/x) x dx + c
= log x - ∫ 1 dx + c
= log x - x + c
∫ x sin x dx = x ∫ sin x dx - ∫x' { ∫ sin x dx}dx + c
= x(-cos x) - ∫ (-cos x)dx + c∫ log x dx = ∫ log x.1 dx
Also Read How To Understand Binary operations , Relations And Functions , Commutative and Associative
Integration Exponential Function
1 ∫ ex dx = ex + c
2 ∫ ax dx = ax / log a + c
3 ∫ log x dx = x log x - x + c
4 ∫ (1/x ) dx = ln |x | + c
Exponential and Derivative Mixed Formula
∫ ex [ f(x) + f '(x)] dx =ex f(x) + c
∫ ex [ sin x + cos x] dx = ex sin x + c
∫ ex [ sin x + cos x] dx = ex sin x + c
Conclusion
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