INDEFINITE INTEGRATION

a function ф(x) is called a primitive or an anti-derivative of a function f(x) if ф'(x) = f(x).

For a function f(x), the collection of all its primitives is called the indefinite integral of f(x0 and is denoted by ∫ f(x) dx.

∫ f(x) dx = ф(x) + C (where C is a constant)

Here ∫ is the integral sign, f(x) is th integrand, x is the variable of integration and dx is the element of integration or differential of x.

The process of finding an indefinite integral of a given function is called integration of the function.

Problem based on direct use of rules of integration and formula for standard integrals.

WORKING RULE

Simplify the integrand till it becomes an algebraic sum of functions so that the rules of integration and formula for standard integral may be directly used.

Example I = ∫ (2 - 3 Sin x) / Cos²x dx

Solution : I = ∫ (2 / Cos²x) dx - 3 ∫(Sin x) / Cos²x dx

I = 2 ∫ Sec²x dx - 3 ∫ Tan x . Sec x dx

I = 2 Tan x - 3 Sec x + C

Integration By Substitution

If g(x) is a continuously differentiable function, then to solve

∫ f(g(x)) g'(x) dx; we substitute g(x) = t and g'(x) dx will be equal to dt.

Hence the problem is transformed to ∫ f(t) dt

Example I = ∫tan x dx = ∫(sin x/cos x)dx

If f(x) = t, f'(x)dx = dt

cos x = t;

-sin x dx = dt

sin x dx = -dt

I = ∫(sin x/cos x)dx = ∫-dt/t = -log |t|+c = - log|cos x|+C

= log |sec x|+C

Integration of ∫ Sin^m x . Cos^n x dx

if m power of Sin x is odd, put Cos x = t .

If n power of Cos x is odd, put Sin x = t.

If both m, and n are odd then put Sin x = t or Cos x = t.

If both m, and n are even then See the sum of power = (m + n).

( a ) if (m + n ) = +ive even Integer then convert trigonometric equation in linear form and integrate it.

( b ) if (m + n ) = -ive even Integer then put Tan x = t and integrate it.

Integrals of the form [1/(x²±a²)]dx

∫(1/(x²+a²)dx = (1/a)tanˉ¹(x/a) + C

∫(1/(x²-a²)dx = (1/2a)log|(x-a)/(x+a)|+C

Integrals of the form [ ∫1/√(ax²+bx+c)]dx or ∫√(ax²+bx+c)]dx

Make ax² + bx + c as perfect square (like A² + X², A² - X² etc )

Let X = t and integrate it.

Integrals of the form [(px+q)/(ax²+bx+c)]dx

Express numerator as

px + q = λ(derivative of denominator) + µ = λ(2ax+b)+µ

Integrals of the functional form [1/(a sin²x + b cos²x +c)]dx

Divide numerator and denominator by cos²x

Replace sec²x by (1 + tan² x)

Put tan x = t

dt = sec²xdx

The integral reduces to ∫[1/(At² +Bt +C)]dt

Integrals of [(a sin x + b cos x +c)/(p sin x + q cos x +r)] dx

Express the numerator as

λ (denominator) + µ (Differential of denominator) + υ

The solution will come as λx + µ log |denominator| + υ ∫ dx / (p sin x + q cos x + r)

Integration by parts

∫ u v dx = u ( ∫ v dx ) - ∫ [ du / dx ∫ v dx ] dx

Integrals of e^ax sin bx dx , e^ax cos bx dx

∫e^ax sinbx dx = [e^ax/(a²+b²)[[a sin bx - b cos bx) +C

∫e^ax cos bx dx = [e^ax/(a²+b²)[[a cos bx + b sin bx) +C

Integrals of √(a² ± x²) and √( x² - a²)

∫√(a²+x²)dx = (1/2) x√(a²+x²) + (1/2) a²log |x+ √(a²+x²)| +C

∫√(a²-x²) = (1/2) x√(a²-x²) + (1/2) a²sin^-1(x/a) + C

∫√(x²-a²) = (1/2) x√(x²-a²)- (1/2) a²log | x+ √ ( x² - a² ) | +C

Integration of Function [G(x)/(P√Q)]dx

∫ [ G(x) /(P√Q)] dx

When both P and Q are linear functions of x put Q = t²

For more specific functional form

P√Q = (ax+b)√(cx+d)

Hence put cx+d = t²

Integration of [( x² + 1 ) / ( x4+ λ x²+1 ) ] dx

Divide the numerator and denominator by x² and put x + 1/x at t or x - 1/x as t as required