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Where do I find the Integration of a particular function Formula?

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  • Students can find the formula of integration for a particular function in this article. It is important to memorise these formulas to solve the integral problem easily. Below is the integration of a particular function formula.

    The formula of a particular function:

    1. Power Rule: 

    ∫ x n d x = x n + 1 n + 1 + C ( n ≠ − 1 )

    2. Reciprocal Function:

    ∫ 1 x d x = ln | x | + C

    3. Trigonometric Function: 

    • ∫ sin x d x = − cos x + C
    • ∫ cos x d x = sin x + C
    • ∫ sec 2 x d x = tan x + C
    • ∫ csc 2 x d x = − cot x + C
    • ∫ sec x tan x d x = sec x + C
    • ∫ csc x cot x d x = − csc x + C

    4. Inverse Trigonometric Function

    • ∫ 1 1 + x 2 d x = tan − 1 x + C
    • ∫ 1 1 − x 2 d x = sin − 1 x + C
    • ∫ − 1 1 − x 2 d x = cos − 1 x + C

Similar Questions for you

How to remember integrals of some particular functions?
V
Vikash Kumar Vishwakarma

Below are a few important tips to remember the integrals of some particular functions:
1. Know the derivatives for each integral.
2. Make yourself familiar with the standard formulas first.
3. Practice daily for better memory.
4. Group similar formulas

∫ ( s i n x − c o s x ) s i n 2 x s i n x c o s 2 x + t a n x s i n 3 x d x is equal to
A
alok kumar singh

∫ ( s i n x − c o s x ) s i n 2 x t a n x ( s i n 3 x + c o s 3 x ) d x

∫ ( s i n x − c o s x ) s i n x c o s x s i n 3 x + c o s 3 x d x , put sin3x + cos3x = t(3 sin2x×cosx – 3cos2xsinx) dx = dt

-> 1 3 ∫ d t t

= l n t 3 + c

= l n | s i n 3 x + c o s 3 x | 3 + c

             

           

l i m x → π − 2 ∫ x 3 ( π 2 ) 2 c o s t 1 / 3 d t ( x − π 2 ) 2

A
alok kumar singh

l i m h → 0 ∫ ( π 2 − h ) ( π 2 ) 3 c o s ( t 1 / 3 ) d t h 2

= l i m h → 0 0 + 3 ( π 2 − h ) 2 c o s ( π 2 − h ) 2 h

= l i m h → 0 3 ( π 2 − h ) 2 s i n h 2 h

= 3 π 2 8

∫ − π 2 π 2 8 2 c o s x ( 1 + e s i n x ) ( 1 + s i n 4 x ) dx = ap + b log ( 3 + 2 2 )  
A
alok kumar singh

∫ − π 2 π 2 8 2 c o s x ( 1 + e s i n x ) ( 1 + s i n 4 x ) dx

= ∫ 0 π 2 { ( 8 2 c o s x ( 1 + e s i n x ) ( 1 + s i n 4 x ) + 8 2 c o s x ( 1 + e − s i n x ) ( 1 + s i n 4 x ) ) } d x            

= 8 2 ∫ 0 π 2 c o s x 1 + s i n 4 x d x            

Let sin x = t

I = 8 2 ∫ 0 1 d t 1 + t 4            

= 4 2 ∫ 0 1 ( 1 + 1 t 2 ) − ( 1 − 1 t 2 ) t 2 + 1 t 2 d t       

= 4 2 ∫ 0 1 ( 1 + 1 t 2 ) d t ( t − 1 t ) 2 + 2 − 4 2 ∫ 0 1 ( 1 − 1 t 2 ) d t ( t + 1 t ) 2 − 2            

= 4 2 ⋅ 1 2 ( t a n − 1 t − 1 t 2 ) 0 1 − 4 2 ⋅ 1 2 2 [ l o g | t + 1 t − 2 t + 1 t + 2 | ] 0 1         

= 2 π − 2 l o g | 2 − 2 2 + 2 |        

= 2 π + 2 l o g ( 3 + 2 2 )           

a = b = 2

The value of ∫ 0 π / 4 x d x s i n 4 ( 2 x ) + c o s 4 ( 2 x )  equal to
A
alok kumar singh

I = ∫ 0 π / 4 x d x s i n 4 ( 2 x ) + c o s 4 ( 2 x )

           Let 2x = t then   d x = 1 2 d t

I = ∫ t 2 ⋅ 1 2 d t s i n 4 t + c o s 4 t

= 1 4 ∫ 0 π / 2 t   d t s i n 4 t + c o s 4 t d t            

I = 1 4 ∫ 0 π / 2 ( π 2 − t ) d t s i n 4 t + c o s 4 t d t

2 I = 1 4 ∫ 0 π / 2 π 2 d t s i n 4 t + c o s 4 t

2 I = 1 4 ∫ 0 π / 2 π 2 d t s i n 4 t + c o s 4 t

2 I = π 8 ∫ 0 π / 2 s i n 4 t   d t t a n 4 t + 1            

Let tan t = y then

2 I = π 8 ∫ 0 ∞ ( 1 + y 2 ) d y 1 + y 4             

= π 8 ∫ 0 ∞ 1 + 1 y 2 y 2 + 1 y 2 − 2 + 2 d y

= π 8 ∫ 0 ∞ ( 1 + 1 y 2 ) d y 2 + ( y − 1 y ) 2             

Let y−1y=u  

2 I = π 8 ∫ − ∞ ∞ d u 2 + u 2

= π 8 2 [ t a n − 1 4 2 ] − ∞ ∞                  

I = π 2 1 6 2

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