Question

How to derive basic formulas of integration?

Answer 1

The notation, which we're stuck with for historical reasons, is as peculiar as the notation for derivatives: the integral of a function f(x)f(x) with respect to xx is written as

∫f(x)dx∫f(x)dx

The remark that integration is (almost) an inverse to the operation of differentiation means that if

ddxf(x)=g(x)ddxf(x)=g(x)then∫g(x)dx=f(x)+C∫g(x)dx=f(x)+CThe extra CC, called the constant of integration, is really necessary, since after all differentiation kills off constants, which is why integration and differentiation are not exactly inverse operations of each other.

Answer 2

Basically integration is inverse of differentiation .
example :-

if any function y = f(x) + C
where C is constant.
then,
differentiation wrt x
dy/dx = f'(x)
and
integration ,
dy = f'(x).dx
y = f(x) + C

Whole math contains in ILATE
I - inverse function
L - logarithmic function
A - algebraic function
T - Trigonometric function
E- exponential function

only these are the whole mathematics .by the way some extra function also available named as piece - wise function (e.g greatest integer , least integer , fractional part , modulus ) but this can be solved by given condition , in question .

You can derrive basic formula of integration when , you have knowledge about differentiation .

★★ TRIGONOMETRIC FUNCTION★★
we know ,
differentiation of
sinx = cosx
cosx = - sinx
tanx = sec²x
secx = secx.tanx
cosecx = - cosecx.cotx
cotx = - cosec²x

now , I said " integration is inverse of differentiation " .
hence, if we integrate cosx we get sinx
becoz differentiation of sinx = cosx
similarly ,
integration of
sinx = - cosx + C
sec²x = tanx + C
cosec²x = - cotx + C
secx.tanx = secx + C
cossecx.cotx = - cosecx + C

★★ ALGEBRIAC FUNCTIONS ★★
we know , f(x ) = xⁿ is a general and simple algebriac function .
we know , differentiation of xⁿ = n.x^( n -1)

hence, integration of n.x^( n -1) = n.x^{ n -1+ 1}/(n -1 + 1) = x^n

it means integration of xⁿ = x^( n +1)/(n +1)

★★ LOGARITHMIC FUNCTIONS ★★
y = logx
dy/dx = 1/x { differentiation of logx = 1/x
dy = dx/x

hence, integration of 1/x = logx

★★ EXPONENTIAL FUNCTIONS★★
y = a^x = e^xlna
dy/dx =e^lna × lna = a^x.lna

dy = a^x.dx = a^x/lna

★★ INVERSE FUNCTIONS★★
y = sin-¹x
dy/dx =1/√( 1- x²)

so, integration of 1/√( 1-x²) = sin-¹(x )

similarly ,
1/( 1 + x²) = tan-¹x
1/x√( 1 - x²) = sec-¹x


I hope you understand how to relate any function with integration and differentiation and how to find basic formula .