INDEFINITE INTEGRAL
HELLO ASPIRANTS!!!
From now onwards we are going to learn all about calculus. In this series we begin with indefinite integral.
Make sure you have a formula notebook where you write all formula which are essential for you.It is very beneficial for you and it will also help you to revise on last time and due to this you don't have to carry a heavy book and revise every single page.
But before starting I would like to inform you about previous lessons we have already completed whole series on Trigonometry and noun. So if you are not aware about that so click below and check right now.
Edited by :- Indian defence era
So lets begin with all your josh!!!
•∫xn dx=x^(n+1) / (n+1) +c
•∫(1/xn dx)= 1/-(n–1)x^(n-1)
•∫1/x dx= log x +c
•∫ex dx = ex + C
Logarithmic function:-
∫sin x dx = -cos x + c
∫cos x dx = sin x + c
∫cosec²x = -cot x + c
∫ secx tanx dx = sec x + c
∫cosecx cotx dx = - cosec x + c
∫Tan x dx = log |secx + tanx |+ c
∫Tan x dx = - log |cos x| + c
∫cot x dx = log |sin x | + c
∫sec x dx = log | secx + tanx | + c
∫ sec x = log | tan ( π /4 + x/2 )| + c
∫cosec x dx = log |cosec x - cot x | + c
∫cosec x dx = log| tan (x/2)| + c
INVERSE TRIGONOMETRIC
∫sin⁻¹ x dx =( x sin⁻¹ x + √1-x² )+ c
∫cos⁻¹ x dx = ( x cos ⁻¹x -√1-x²)+c
∫tan⁻¹ x dx = x tan⁻¹ x- 1/2 log(1+x²)+c
∫cot⁻¹x dx = x cot⁻¹x +1/2 log (1+x²)+c
∫sec⁻¹ x dx=x sec⁻¹x-cosh⁻¹x+c
∫cosec⁻¹x dx =x cosec⁻¹x +cosh⁻¹x +c
HYPERBOLIC FUNCTION
∫sinhx dx =cosh x + c
∫cosh x dx =sinh x +c
∫sech² x dx = tanh x +c
∫cosech²x = -coth x+c
∫tanh x dx = log |cosh x|+c
∫coth x dx =log|sinh x|+c
∫sech x tanh x dx = -sech x + c
∫cosech x coth x= -cosech x +c
∫sech x dx= sinh⁻¹(tan x)+c=2tan⁻¹(eˣ)
∫cosech x dx =logtanh(x/2)+c=cosh⁻¹(eˣ)
RATIONAL FUNCTION
∫1/(x²-a²)=1/2a log|x-a/x+a|
∫1/(a²-x²)=1/2a log|a+x/a-x|
∫1/(x²+a²)=1/a tan⁻¹ (x/a)
IRRATIONAL FUNCTIONS :-
∫1/√a²-x² = sin⁻¹(x/a)
∫1/√x²-a²=cosh⁻¹(x/a)=log|x+√x²-a²|
∫1/√x²+a²=sinh⁻¹(x/a)=log|x+√x²+a²|
∫√a²-x²=x/2 (√a²-x²)+a²/2 (sin⁻¹x/2)
∫√x²-a²=x/2( √x²-a²)+a²/2 (cos⁻¹ x/a)
∫√x²-a²=x/2 √x²-a² -a²/2 log |x+√x²-a²|
∫√x²+a² =x/2 (√x²+a² )+a²/2( sinh⁻¹x/a)
∫√x²+a² =x/2 √x²-a²+a²/2 log|x+√x²-a²|
∫1/√1-x² =sin⁻¹x +c
∫1/|x|√x²-1=sec⁻¹x+c
some basic fromulas
∫uv dx = u∫v dx - ∫{(du/dx)∫v dx}dx
∫eˣ{f(x)+f'(x)}dx=eˣf(x) +c
∫eᵏˣ{K f(x)+f'(x)}dx=eᵏˣ f(x)+c
∫{f(x)}ⁿ f'(x)dx ={f(x)}ⁿ⁺¹/(n+1)
∫f'(x)/f(x)=log |f(x)|+c
∫xf'(x)+f(x)}dx=x f(x) +c
∫f'(x)/1+{f(x)}² dx = tan⁻¹f(x) + c
∫f'(x)/√f(x) dx =2 √f(x) +c
∫eᵃˣsin(bx+c)dx =eᵃˣ/(a²+b²){a sin(bx+c)-b cos(bx+c)}
∫eᵃˣ cos(bx+c) dx =eᵃˣ/(a²+b²){a cos(bx+c)+b sin(bx+c) dx
SOME IMPORTANT SUBSITUTION
√a²-x² :- x=a sinθ
√a²+x² :- x=a tanθ
√x²-a² :- x=a secθ
√a-x/a+x :-x=a cos2θ
√(x-α)(x-β) :- x=αcos²θ+βsin²θ
For queries and inquiry ask me