Mathematics INTEGRATION BY SUBSTITUTION ,BY PARTS AND BY PARTIAL FRACTIONS

SUBSTITUTION :

Often it is not possible to convert an integral into loving integral just by simple manipulation. Then required some techniques to convert an integral into loving integral. This techniques are following.

`text(SUBSTITUTION :)`

Theory : `I= int f(x) dx ` and let `x = phi (z)`

`(dI)/dx =f(x)`; `dx/dz = phi ' (z)`

`=> (dI)/dz = ((dI)/dx )*dx/dz = f(x) * phi ' (z)` or `(dI)/dz = f (phi (z)) phi'(z)`

Hence `I= int f (phi (z)) phi' (z) dz` ....................(1)

Substitution is said to be appropriate if the integrand in (1) is a loving one. (standard integral)
If `int [f(x)]^n f' (x) dx ` or `int (f'(x)) /[f(x)]^n dx`

start with `f(x) =t`

`int (tan x)dx = ln sec x+C = -lm(cos x) +C`;
`int (cot x)dx =ln (sin x)` (loving integrals)

`text(Proof:)` `int tan xdx = int sinx/cosx dx`
put `cos x =t` to get `int -dt/t= -ln t+c = -ln (cos x)+c =ln (sec x)+c`

`text(General Substitution )`:

`(tt( (sqrt (a^2-x^2) ; x=a sin theta) , (sqrt (a^2 +x^2) ; x = a tan theta) ,(sqrt(x^2 -a^2) ; x = a sec theta) , (sqrt ((a^2-x^2)/(a^2+x^2)); x^2 =a^2 cos 2 theta) )]`

note that `int sqrt (a^2 +x^2) dx ` & `int sqrt (x^2-a^2)dx`

to be executed by parts.

`int dx/sqrt(x^2+a^2)= ln (x+ sqrt (x^2+a^2))` & `int dx/sqrt(x^2-a^2)=ln (x+ sqrt (x^2 -a^2))` (loving integrals)

`text(Loving Integrals:- )`

`int dx/sqrt(a^2-x^2) =sin^-1 (x/a)+c`; `quadquadquadquadquadquadint dx/(x^2+a^2) =1/a tan^-1(x/a)+c`;

`int dx/( root x (x^2-a^2))=1/a sec^-1 (x/a)+c`; `quadquadquadquadquadquadint dx/sqrt(x^2 -a^2)=ln (x+ sqrt(x^2 -a))+c`;

`int dx / sqrt (x^2 +a^2) = ln (x + sqrt (x^2-a^2))+c`

Integration By Parts :

If `f(x)` and `g (x)` are derivable functions then
`d/dx [f(x)* g(x) ]= f(x) * g' (x) +g(x) * f'(x)`

`:.` `int f(x) *g'(x) dx =f(x) *g(x) -int g(x) * f'(x) dx`

`I = int underbrace (f(x))_(I) * underbrace (g(x))_(II) dx`

`= 1^(st)` function x integral of `2^(nd) -``int (`diff. co-eff. of `1^(st)) xx (` integral of `2^(nd)) dx`

Remember `text(ILATE)` for deciding the choice of the first and second function which is not arbitrary.
Here `quadquadquadquadtext( I)` for inverse trigonometric function
`quadquadquadquadquadquadquadquadtext( L)` for Logarithmic function
`quadquadquadquadquadquadquadquadquadtext(A)` for Algebraic function
`quadquadquadquadquadquadquadtext( T)` for Trigonometric function
`quadquadquadquadquadquadquadquad text (E)` for Exponential Function

`text(Two Classic Integrands : )`

`(a)` `inte^x (f(x) +f'(x))dx =e^x f(x) +Cquadquadquad` & `quadquadquad (b)` `int (f(x) + x f' (x))dx =x f(x) +C`

`text(Proof:)`
`(a)` `int e^x (f(x) +f' (x))dx = int e^x f(x) dx + int underbrace (e^x)_(I) underbrace(f'(x) )_ (II)dx`

`= int e^x f(x) dx +e^x f(x)- int e^x f(x) dx +c = e^x f(x) +c`

`(b)` `int f(x) dx + x f'(x) dx =int f(x) dx+ int underbrace(x)_(I) underbrace(f' (x))_(II) dx`

`= int f (x) dx +x f(x) - int f(x) dx = x f(x) +c`

Partial Fraction :

This technique is used if a rational function is being integrated whose denomenator can be is factorised. If degree of numerator is greater then degree of denomenator then first devide numerator by denomenator.

Loving lntegrands `int dx /(a^2-x^2) =1/2a ln ((a+x)/(a-x))` & `int dx/ (x^2 -a^2) = 1/2a ln ((x-a)/(x+a))`

`text(Integrals Of Trigonometric Functions :)`

`Type -1 : ` `int dx/(a+bsin^2 x)` , `int dx/(a+bcos^2 x)` , `int dx/ (a sin^2 x+ b cos^2 x + c sin x cosx)` , `dx/ (a cos x + b sin x)^2`

`Type- 2: ``int dx/(a+ b sin x)`, `int dx/ (a+ b cos x)` , `int dx/(a +bsin x + cosx)`

Convert `sin x` and `cos x` into their corresponding tangent to half the angles and

put `tan (x/2) =t`

`Type-3: ` `int (a sin x + b cos x +c)/(l sin x + m cos x+n)dx`; `N^r= A(D^r)+B (d/dx D^r)+C`

`Type-4: ``int (x^2+1)/(x^4 +kx^2 +1)dx` or `int (x^2 -1)/(x^4 +kx^2 +1)dx`

Divide `N^r` and `D^r` by `x^2` and take suitable substitution.