4.2.9
Jinteграls mein a 2 − x 2 a 2 + x 2 x 2 − a 2 ugly lagte hain — square root ki wajah se. Hum trig functions kyun substitute karte hain? Kyunki Pythagorean identities ek sum/difference of squares ko ek single squared term mein badal deti hain, toh root cleanly gayab ho jaata hai.
1 − sin 2 θ = cos 2 θ ⇒ a 2 − a 2 sin 2 θ = a cos θ 1 + tan 2 θ = sec 2 θ ⇒ a 2 + a 2 tan 2 θ = a sec θ sec 2 θ − 1 = tan 2 θ ⇒ a 2 sec 2 θ − a 2 = a tan θ
Hum ek algebraic root ko ek aisi trig integral se trade karte hain jise hum pehle se solve karna jaante hain.
Definition Pattern → substitution table
Integrand mein radical
Substitution
d x Root simplify hokar
Range of θ
a 2 − x 2 x = a sin θ d x = a cos θ d θ a cos θ [ − 2 π , 2 π ]
a 2 + x 2 x = a tan θ d x = a sec 2 θ d θ a sec θ ( − 2 π , 2 π )
x 2 − a 2 x = a sec θ d x = a sec θ tan θ d θ a tan θ [ 0 , 2 π ) ∪ [ π , 2 3 π )
Yahan a > 0 Range restriction guarantee karta hai ki root non-negative ho, isliye ⋅ = +
Step 1 — substitute karo. Maano x = a sin θ d x = a cos θ d θ Yeh step kyun? a 2 − x 2 1 − sin 2 θ
Step 2 — root simplify karo.
a 2 − x 2 = a 2 − a 2 sin 2 θ = a 2 cos 2 θ = a cos θ . Yeh + a cos θ ± Kyunki θ ∈ [ − 2 π , 2 π ] cos θ ≥ 0
Step 3 — assemble karo.
∫ a 2 − x 2 d x = ∫ a cos θ ⋅ a cos θ d θ = a 2 ∫ cos 2 θ d θ .
Step 4 — power-reduction. cos 2 θ = 2 1 + c o s 2 θ = 2 a 2 ( θ + 2 s i n 2 θ ) + C = 2 a 2 θ + 2 a 2 sin θ cos θ + C . Yeh step kyun? cos 2
Step 5 — back-substitute karo ek right triangle use karke: sin θ = a x ⇒ θ = arcsin a x cos θ = a a 2 − x 2 ∫ a 2 − x 2 d x = 2 a 2 arcsin a x + 2 x a 2 − x 2 + C
∫ 9 − x 2 d x Step 1. a = 3 x = 3 sin θ d x = 3 cos θ d θ Kyun? Root a 2 − x 2 Step 2. 9 − x 2 = 3 cos θ = ∫ 3 cos θ 3 cos θ d θ = ∫ d θ = θ + C . Yeh step kyun? Substitution ne sab kuch cancel kar diya — yahi toh fayda hai.
Step 3. θ = arcsin 3 x = arcsin 3 x + C .
∫ x 2 x 2 + 4 d x Step 1. Root x 2 + 4 ⇒ a = 2 x = 2 tan θ d x = 2 sec 2 θ d θ Step 2. x 2 + 4 = 2 sec θ x 2 = 4 tan 2 θ ∫ 4 t a n 2 θ ⋅ 2 s e c θ 2 s e c 2 θ d θ = 4 1 ∫ t a n 2 θ s e c θ d θ . Step 3. sin/cos mein likhte hain: t a n 2 θ s e c θ = s i n 2 θ / c o s 2 θ 1/ c o s θ = s i n 2 θ c o s θ . Kyun? Ab ek clean u Step 4. Maano u = sin θ 4 1 ∫ u 2 d u = − 4 u 1 + C = − 4 s i n θ 1 + C . Step 5. Triangle: tan θ = 2 x = x = 2 = x 2 + 4 sin θ = x 2 + 4 x = − 4 x x 2 + 4 + C .
∫ x x 2 − 1 d x Step 1. a = 1 x = sec θ d x = sec θ tan θ d θ Step 2. x 2 − 1 = tan θ = ∫ sec θ tan θ sec θ tan θ d θ = ∫ tan 2 θ d θ . Step 3. tan 2 θ = sec 2 θ − 1 kyun? directly integrable hai): = tan θ − θ + C . Step 4. Wapas: tan θ = x 2 − 1 θ = arcsec x = arccos x 1 = x 2 − 1 − arcsec x + C .
Recall Dekhne se pehle substitution predict karoHar ek ke liye, sub ka naam batao, phir neeche check karo.
∫ ( 4 − x 2 ) 3/2 d x forecast: x = 2 sin θ ∫ x 2 − 25 x 2 d x forecast: x = 5 sec θ ∫ ( x 2 + 7 ) 2 d x forecast: x = 7 tan θ
Sab sahi hain — radical (ya same sign structure wala ( ) 2
d x Kyun sahi lagta hai: tum mechanically x d x Fix: d x x = a tan θ ⇒ d x = a sec 2 θ d θ
a 2 cos 2 θ = a cos θ Kyun sahi lagta hai: u 2 = u Fix: u 2 = ∣ u ∣ range restriction precisely isliye hai ki cos θ ≥ 0 tan θ ≥ 0 ∣ ⋅ ∣
θ Kyun sahi lagta hai: lagta hai integral "solve" ho gaya.
Fix: Substitution se reference right triangle banao aur sin θ , cos θ , tan θ , θ x
x 2 − a 2 a sin θ Kyun sahi lagta hai: sine pehli sub hai jo seekhi thi.
Fix: Sign structure match karo: pehle a 2 a 2 − x 2 a 2 + x 2 x 2 − a 2
"SIN a x
Short mein: a 2 − x 2 → sin a 2 + x 2 → tan x 2 − a 2 → sec S–T–S , − , + , −
Recall Feynman: ek 12-saal ke bacche ko samjhaoSocho ek ladder wall se lagi hai. Ladder ki length, wall ki height, aur floor ka distance hamesha ek right triangle banate hain, aur woh Pythagoras se lock hain. a 2 − x 2 maante hain ki x θ . Jab sab kuch angles mein ho jaata hai, square root gayab ho jaata hai (triangles hamesha nicely close hote hain), hum aasaan angle problem solve karte hain, phir angle ko triangle use karke original side length mein wapas convert kar dete hain. Humne ek mushkil problem ko triangle ke bhes mein chhupa diya, solve kiya, aur phir bhes utaar diya.
a 2 − x 2 x = a sin θ d x = a cos θ d θ → a cos θ
a 2 + x 2 x = a tan θ d x = a sec 2 θ d θ → a sec θ
x 2 − a 2 x = a sec θ d x = a sec θ tan θ d θ → a tan θ
Trig sub ke neeche root kyun simplify hoti hai? Pythagorean identities sum/difference of squares ko ek single squared term mein convert karti hain.
a 2 cos 2 θ = a cos θ ± Chosen θ cos θ ≥ 0 ∣ cos θ ∣ = cos θ
∫ a 2 − x 2 d x = ? arcsin a x + C
∫ a 2 + x 2 d x = ? a 1 arctan a x + C
∫ a 2 − x 2 d x = ? 2 a 2 arcsin a x + 2 x a 2 − x 2 + C .
Kisi bhi trig sub ka last mandatory step kya hai? Reference right triangle use karke x
∫ tan 2 θ d θ = ? tan θ − θ + C tan 2 = sec 2 − 1
Pythagorean identities — woh engine jo root khatam karta hai.Integration by substitution (u-sub) — trig sub ek clever change of variable hai.Power-reduction & double-angle formulas — ∫ cos 2 θ , sin 2 θ Reference right triangle method — back-substitution ke liye.Partial fractions — kuch rational integrals ke liye alternative.Arc length and surface area — 1 + x 2 Hyperbolic substitution — x = a sinh u a cosh u
Right triangle back-substitution
Closed-form antiderivative
