📦 Area aur Integration
Socho ek curved field hai jisme tum crop ugaana chahte ho. Straight lines se area calculate karna mushkil hai. Lekin agar tum field ko infinitely thin strips mein baat lo — har strip ek rectangle jaisi lagegi — aur sab rectangles ka area add karo — toh total area milega! Yahi integration ka idea hai. Formally: ∫ f(x)dx = limit of sum of infinitely many infinitely thin rectangles.
1. Standard Integration Formulas
∫xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ -1)
∫1/x dx = ln|x| + C
∫eˣ dx = eˣ + C
∫aˣ dx = aˣ/ln(a) + C
∫sin x dx = -cos x + C
∫cos x dx = sin x + C
∫sec²x dx = tan x + C
∫cosec²x dx = -cot x + C
∫sec x tan x dx = sec x + C
∫1/√(1-x²) dx = sin⁻¹x + C
∫1/(1+x²) dx = tan⁻¹x + C
2. Integration by Substitution
Jab integrand mein ek function aur uski derivative dono hain — substitution karo!
Put t = g(x), then dt = g'(x)dx → ∫f(g(x))g'(x)dx = ∫f(t)dt
✅ ∫2x·cos(x²) dx
t = x², dt = 2x dx
∫cos(t) dt = sin(t) + C = sin(x²) + C
3. Integration by Parts
∫u·v dx = u·∫v dx - ∫(du/dx · ∫v dx) dx
ILATE Rule — choose u in this order:
I — Inverse trig
L — Logarithmic
A — Algebraic (polynomial)
T — Trigonometric
E — Exponential
✅ ∫x·eˣ dx (ILATE: u=x, v=eˣ)
= x·eˣ - ∫eˣ dx = x·eˣ - eˣ + C = eˣ(x-1) + C
✅ ∫x·sin x dx
u=x, v=sin x → = x(-cos x) - ∫(-cos x)dx = -x cos x + sin x + C
4. Integration by Partial Fractions
Jab integrand ek rational function ho (polynomial/polynomial) — partial fractions use karo.
/* Types */
(px+q)/[(x-a)(x-b)] = A/(x-a) + B/(x-b)
(px²+qx+r)/[(x-a)(x-b)(x-c)] = A/(x-a) + B/(x-b) + C/(x-c)
(px+q)/[(x-a)²] = A/(x-a) + B/(x-a)²
5. Definite Integrals
∫[a to b] f(x) dx = [F(x)]ₐᵇ = F(b) - F(a)
where F'(x) = f(x)
Fundamental Theorem: d/dx ∫[a to x] f(t)dt = f(x)
Key Properties:
∫[a to b]f(x)dx = -∫[b to a]f(x)dx
∫[a to a]f(x)dx = 0
∫[0 to 2a]f(x)dx = 2∫[0 to a]f(x)dx if f(2a-x)=f(x)
∫[-a to a]f(x)dx = 2∫[0 to a]f(x)dx if f is even
= 0 if f is odd
✅ Evaluate ∫[0 to π] sin x dx
= [-cos x]₀^π = -cos(π) - (-cos 0) = 1 + 1 = 2
6. Important Standard Integrals to Remember
∫dx/(x²-a²) = (1/2a)ln|(x-a)/(x+a)| + C
∫dx/(a²-x²) = (1/2a)ln|(a+x)/(a-x)| + C
∫dx/√(x²-a²) = ln|x+√(x²-a²)| + C
∫dx/√(a²-x²) = sin⁻¹(x/a) + C
∫√(a²-x²)dx = (x/2)√(a²-x²) + (a²/2)sin⁻¹(x/a) + C
7. Practice Questions
Q1. ∫(x+1)/(x²+2x+3) dx
Solution
Numerator = ½ × d/dx(x²+2x+3) = ½(2x+2). So integral = (½)ln(x²+2x+3) + C
Q2. ∫x²·ln x dx
Solution
IBP: u=ln x, v=x². = (x³/3)ln x - ∫(x³/3)(1/x)dx = (x³/3)ln x - x³/9 + C
Q3. Evaluate ∫[0 to 1] x/(1+x²) dx
Solution
t=1+x², dt=2x dx. = ½∫[1 to 2] (1/t)dt = ½[ln t]₁² = ½ ln 2